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Definite Integration — JEE Previous Year Questions

Every Definite Integration question asked in JEE Main and JEE Advanced across the last 186 papers — 267 questions, each with its correct answer. Free to read, no account needed.

Questions

267

Papers it appeared in

168/186

Appearance rate

90%

All 267 Definite Integration questions

Most recent papers first.

Q1·MathematicsNumericalJEE Main 2026
If ∫π/6π/4(cot⁡(x−π3)cot⁡(x+π3)+1)dx=αlog⁡e(3−1)\displaystyle\int_{\pi/6}^{\pi/4} \left(\cot\left(x - \dfrac{\pi}{3}\right)\cot\left(x + \dfrac{\pi}{3}\right) + 1\right)dx = \alpha \log_e(\sqrt{3} - 1)∫π/6π/4​(cot(x−3π​)cot(x+3π​)+1)dx=αloge​(3​−1), then 9α29\alpha^29α2 is equal to ________.

Correct answer: 12

Step-by-step solution →
Q2·MathematicsSingle correctJEE Main 2026
The value of the integral ∫02x(x2+x+1)(x+1)(x4+x2+1) dx\displaystyle\int_{0}^{2} \dfrac{\sqrt{x(x^2+x+1)}}{(\sqrt{x+1})(\sqrt{x^4+x^2+1})}\,dx∫02​(x+1​)(x4+x2+1​)x(x2+x+1)​​dx is equal to:
  1. (A)13log⁡e(3−22)\dfrac{1}{3}\log_e(3 - 2\sqrt{2})31​loge​(3−22​)
  2. (B)23log⁡e(4+2)\dfrac{2}{3}\log_e(4 + \sqrt{2})32​loge​(4+2​)
  3. (C)23log⁡e(3+22)\dfrac{2}{3}\log_e(3 + 2\sqrt{2})32​loge​(3+22​)
  4. (D)13log⁡e(1+62)\dfrac{1}{3}\log_e(1 + 6\sqrt{2})31​loge​(1+62​)

Correct answer: (C)

Step-by-step solution →
Q3·MathematicsSingle correctJEE Main 2026
The value of the integral ∫−11(x3+∣x∣+1x2+2∣x∣+1)dx\int_{-1}^{1} \left( \frac{x^{3} + |x| + 1}{x^{2} + 2|x| + 1} \right) dx∫−11​(x2+2∣x∣+1x3+∣x∣+1​)dx is equal to :
  1. (A)3log⁡e23\log_{e} 23loge​2
  2. (B)2log⁡e22\log_{e} 22loge​2
  3. (C)5log⁡e35\log_{e} 35loge​3
  4. (D)3log⁡e33\log_{e} 33loge​3

Correct answer: (B)

Step-by-step solution →
Q4·MathematicsSingle correctJEE Main 2026
The value of the integral ∫−π/4π/4(32cos⁡4x1+esin⁡x)dx\int_{-\pi/4}^{\pi/4}\left(\frac{32\cos^4 x}{1+e^{\sin x}}\right)dx∫−π/4π/4​(1+esinx32cos4x​)dx is:
  1. (A)4π + 2
  2. (B)3π + 8
  3. (C)3π + 4
  4. (D)4π + 3

Correct answer: (B)

Step-by-step solution →
Q5·MathematicsSingle correctJEE Main 2026
Let A=[13−121α01−1]A = \begin{bmatrix} 1 & 3 & -1 \\ 2 & 1 & \alpha \\ 0 & 1 & -1 \end{bmatrix}A=​120​311​−1α−1​​ be a singular matrix. Let f(x)=∫0x(t2+2t+3) dtf(x) = \int_{0}^{x} (t^{2} + 2t + 3)\, dtf(x)=∫0x​(t2+2t+3)dt, x∈[1,α]x \in [1, \alpha]x∈[1,α]. If MMM and mmm are respectively the maximum and the minimum values of fff in [1,α][1, \alpha][1,α], then 3(M−m)3(M - m)3(M−m) is equal to :
  1. (A)64
  2. (B)68
  3. (C)72
  4. (D)76

Correct answer: (B)

Step-by-step solution →
Q6·MathematicsSingle correctJEE Main 2026
The value of the integral ∫π/6π/3(4−csc⁡2xcos⁡4x)dx\int_{\pi/6}^{\pi/3} \left( \frac{4 - \csc^2 x}{\cos^4 x} \right) dx∫π/6π/3​(cos4x4−csc2x​)dx is:
  1. (A)113\frac{11}{\sqrt{3}}3​11​
  2. (B)163\frac{16}{\sqrt{3}}3​16​
  3. (C)3233\frac{32}{3\sqrt{3}}33​32​
  4. (D)6433\frac{64}{3\sqrt{3}}33​64​

Correct answer: (C)

Step-by-step solution →
Q7·MathematicsSingle correctJEE Main 2026
Let (21−a+21+a)(2^{1-a} + 2^{1+a})(21−a+21+a), f(a), (3a+3−a)(3^{a} + 3^{-a})(3a+3−a) be in A.P. and α be the minimum value of f(a). Then the value of the integral ∫log⁡e(α−1)log⁡e(α)dx(e2x−e−2x)\int_{\log_e(\alpha-1)}^{\log_e(\alpha)} \frac{dx}{(e^{2x} - e^{-2x})}∫loge​(α−1)loge​(α)​(e2x−e−2x)dx​ is :
  1. (A)12log⁡e(43)\frac{1}{2}\log_e\left(\frac{4}{3}\right)21​loge​(34​)
  2. (B)14log⁡e(43)\frac{1}{4}\log_e\left(\frac{4}{3}\right)41​loge​(34​)
  3. (C)12log⁡e(85)\frac{1}{2}\log_e\left(\frac{8}{5}\right)21​loge​(58​)
  4. (D)14log⁡e(85)\frac{1}{4}\log_e\left(\frac{8}{5}\right)41​loge​(58​)

Correct answer: (B)

Step-by-step solution →
Q8·MathematicsSingle correctJEE Main 2026
The value of the integral ∫0∞log⁡e(x)x2+4 dx\int_0^\infty \frac{\log_e(x)}{x^2 + 4}\,dx∫0∞​x2+4loge​(x)​dx is:
  1. (A)πlog⁡e(2)2\frac{\pi \log_e(2)}{2}2πloge​(2)​
  2. (B)πlog⁡e(2)4\frac{\pi \log_e(2)}{4}4πloge​(2)​
  3. (C)1+πlog⁡e(2)1 + \pi \log_e(2)1+πloge​(2)
  4. (D)2+πlog⁡e(2)2 + \pi \log_e(2)2+πloge​(2)

Correct answer: (B)

Step-by-step solution →
Q9·MathematicsSingle correctJEE Main 2026
The integral ∫01cot⁡−1(1+x+x2)dx\int_{0}^{1}\cot^{-1}(1 + x + x^{2})dx∫01​cot−1(1+x+x2)dx is equal to:
  1. (A)2tan⁡−12+12log⁡e(54)+π22\tan^{-1}2 + \frac{1}{2}\log_{e}\left(\frac{5}{4}\right) + \frac{\pi}{2}2tan−12+21​loge​(45​)+2π​
  2. (B)2tan⁡−12+12log⁡e(54)−π22\tan^{-1}2 + \frac{1}{2}\log_{e}\left(\frac{5}{4}\right) - \frac{\pi}{2}2tan−12+21​loge​(45​)−2π​
  3. (C)2tan⁡−12−12log⁡e(54)+π22\tan^{-1}2 - \frac{1}{2}\log_{e}\left(\frac{5}{4}\right) + \frac{\pi}{2}2tan−12−21​loge​(45​)+2π​
  4. (D)2tan⁡−12−12log⁡e(54)−π22\tan^{-1}2 - \frac{1}{2}\log_{e}\left(\frac{5}{4}\right) - \frac{\pi}{2}2tan−12−21​loge​(45​)−2π​

Correct answer: (D)

Step-by-step solution →
Q10·MathematicsNumericalJEE Main 2026
Let fff be a twice differentiable function such that f(x)=∫0xtan⁡(t−x)dt−∫0xf(t)tan⁡t dtf(x) = \int_{0}^{x}\tan(t - x)dt - \int_{0}^{x}f(t)\tan t\,dtf(x)=∫0x​tan(t−x)dt−∫0x​f(t)tantdt, x∈(−π2,π2)x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)x∈(−2π​,2π​). Then f′′(π6)+12f′(−π6)+f(π6)f''\left(\frac{\pi}{6}\right) + 12f'\left(-\frac{\pi}{6}\right) + f\left(\frac{\pi}{6}\right)f′′(6π​)+12f′(−6π​)+f(6π​) is equal to ______

Correct answer: 5

Step-by-step solution →
Q11·MathematicsSingle correctJEE Main 2026
Let fff be a real polynomial of degree nnn such that f(x)=f′(x)f′′(x)f(x) = f'(x)f''(x)f(x)=f′(x)f′′(x), for all x∈Rx \in \mathbb{R}x∈R. If f(0)=0f(0) = 0f(0)=0, then 36(f′(2)+f′′(2)+∫02f(x) dx)36\left(f'(2) + f''(2) + \int_0^2 f(x)\, dx\right)36(f′(2)+f′′(2)+∫02​f(x)dx) is equal to:
  1. (A)424242
  2. (B)464646
  3. (C)565656
  4. (D)666666

Correct answer: (C)

Step-by-step solution →
Q12·MathematicsSingle correctJEE Main 2026
Let ∫−22(∣sin⁡x∣+[xsin⁡x])dx=2(3−cos⁡2)+β\int_{-2}^{2}\left(|\sin x| + [x \sin x]\right) dx = 2(3 - \cos 2) + \beta∫−22​(∣sinx∣+[xsinx])dx=2(3−cos2)+β, where [⋅][\cdot][⋅] is the greatest integer function. Then βsin⁡(β2)\beta \sin\left(\frac{\beta}{2}\right)βsin(2β​) equals:
  1. (A)111
  2. (B)222
  3. (C)444
  4. (D)888

Correct answer: (B)

Step-by-step solution →
Q13·MathematicsSingle correctJEE Main 2026
The value of ∫020π(sin⁡4x+cos⁡4x) dx\int_0^{20\pi} (\sin^4 x + \cos^4 x)\, dx∫020π​(sin4x+cos4x)dx is equal to:
  1. (A)15π2\frac{15\pi}{2}215π​
  2. (B)25π25\pi25π
  3. (C)15π15\pi15π
  4. (D)25π2\frac{25\pi}{2}225π​

Correct answer: (C)

Step-by-step solution →
Q14·MathematicsSingle correctJEE Main 2026
Let [⋅][\cdot][⋅] denote the greatest integer function. Then the value of ∫03(ex+e−x[x]!)dx\int_{0}^{3}\left(\frac{e^{x} + e^{-x}}{[x]!}\right)dx∫03​([x]!ex+e−x​)dx is :
  1. (A)e2+e3−1e2−1e3e^{2} + e^{3} - \frac{1}{e^{2}} - \frac{1}{e^{3}}e2+e3−e21​−e31​
  2. (B)12(e2+e3−1e2−1e3)\frac{1}{2}\left(e^{2} + e^{3} - \frac{1}{e^{2}} - \frac{1}{e^{3}}\right)21​(e2+e3−e21​−e31​)
  3. (C)e2+e3−12e2−12e3e^{2} + e^{3} - \frac{1}{2e^{2}} - \frac{1}{2e^{3}}e2+e3−2e21​−2e31​
  4. (D)12(e2+e3)−1e2−1e3\frac{1}{2}\left(e^{2} + e^{3}\right) - \frac{1}{e^{2}} - \frac{1}{e^{3}}21​(e2+e3)−e21​−e31​

Correct answer: (B)

Step-by-step solution →
Q15·MathematicsNumericalJEE Main 2026
If α=∫023log⁡2(x2+4)dx+∫242x−4 dx\alpha = \int_{0}^{2\sqrt{3}}\log_{2}\left(x^{2} + 4\right)dx + \int_{2}^{4}\sqrt{2^{x} - 4}\,dxα=∫023​​log2​(x2+4)dx+∫24​2x−4​dx, then α2\alpha^{2}α2 is equal to ________.

Correct answer: 192

Step-by-step solution →
Q16·MathematicsNumericalJEE Main 2026
The value of ∑r=120(∣π(∫0rx ∣sin⁡πx ∣dx)∣)\sum_{r=1}^{20}\left(\left|\sqrt{\pi\left(\int_{0}^{r} x\,|\sin \pi x\,|dx\right)}\right|\right)∑r=120​(​π(∫0r​x∣sinπx∣dx)​​) is ____.

Correct answer: 210

Step-by-step solution →
Q17·MathematicsSingle correctJEE Main 2026
Let [·] denote the greatest integer function. Then ∫−π2π2(12(3+[x])3+[sin⁡x]+[cos⁡x])dx\displaystyle\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left(\dfrac{12(3+[x])}{3+[\sin x]+[\cos x]}\right) dx−2π​∫2π​​(3+[sinx]+[cosx]12(3+[x])​)dx is equal to:
  1. (A)15π+415\pi + 415π+4
  2. (B)11π+211\pi + 211π+2
  3. (C)13π+113\pi + 113π+1
  4. (D)12π+512\pi + 512π+5

Correct answer: (B)

Step-by-step solution →
Q18·MathematicsSingle correctJEE Main 2026
Let fff be a polynomial function such that f(x2+1)=x4+5x2+2f(x^2 + 1) = x^4 + 5x^2 + 2f(x2+1)=x4+5x2+2, for all x∈Rx \in \mathbb{R}x∈R. Then ∫03f(x) dx\int\limits_{0}^{3} f(x)\,dx0∫3​f(x)dx is equal to
  1. (A)413\dfrac{41}{3}341​
  2. (B)332\dfrac{33}{2}233​
  3. (C)272\dfrac{27}{2}227​
  4. (D)53\dfrac{5}{3}35​

Correct answer: (B)

Step-by-step solution →
Q19·MathematicsNumericalJEE Main 2026
If f(x) satisfies the relation f(x)=ex+∫01(y+xex)f(y) dyf(x) = e^{x} + \int\limits_{0}^{1} (y + xe^{x}) f(y)\, dyf(x)=ex+0∫1​(y+xex)f(y)dy, then e+f(0)e + f(0)e+f(0) is equal to ________.

Correct answer: 2

Step-by-step solution →
Q20·MathematicsNumericalJEE Main 2026
Let a differentiable function fff satisfy the equation ∫036f(tx36)dt=4αf(x).\int\limits_{0}^{36} f\left(\frac{\mathrm{tx}}{36}\right) \mathrm{dt} = 4\alpha f(\mathrm{x}).0∫36​f(36tx​)dt=4αf(x). If y = fff(x) is a standard parabola passing through the points (2, 1) and (–4, β), Then βα\beta^{\alpha}βα is equal to ______.

Correct answer: 64

Step-by-step solution →
Q21·MathematicsNumericalJEE Main 2026
The number of elements in the set S={x:x∈[0,100] and ∫0xt2sin⁡(x−t)dt=x2}S = \left\{ x : x \in [0,100] \text{ and } \int_{0}^{x} t^2 \sin(x-t)dt = x^2 \right\}S={x:x∈[0,100] and ∫0x​t2sin(x−t)dt=x2} is.....

Correct answer: 16

Step-by-step solution →
Q22·MathematicsSingle correctJEE Main 2026
The value of the integral ∫π245π24dx1+tan⁡2x3\int_{\frac{\pi}{24}}^{\frac{5\pi}{24}} \frac{dx}{1 + \sqrt[3]{\tan 2x}}∫24π​245π​​1+3tan2x​dx​ is :
  1. (A)π12\frac{\pi}{12}12π​
  2. (B)π18\frac{\pi}{18}18π​
  3. (C)π6\frac{\pi}{6}6π​
  4. (D)π3\frac{\pi}{3}3π​

Correct answer: (A)

Step-by-step solution →
Q23·MathematicsNumericalJEE Main 2026
Let [•] be the greatest integer function. If α=∫064(x1/3−[x1/3])dx\alpha = \int_0^{64}\left(x^{1/3} - \left[x^{1/3}\right]\right)dxα=∫064​(x1/3−[x1/3])dx, then 1π∫0απ(sin⁡2θsin⁡6θ+cos⁡6θ)dθ\frac{1}{\pi}\int_0^{\alpha\pi}\left(\frac{\sin^2\theta}{\sin^6\theta + \cos^6\theta}\right)d\thetaπ1​∫0απ​(sin6θ+cos6θsin2θ​)dθ is equal to ____.

Correct answer: 36

Step-by-step solution →
Q24·MathematicsSingle correctJEE Main 2026
The value of ∫−π2π2(1[x]+4)dx\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left(\frac{1}{[x]+4}\right)dx∫−2π​2π​​([x]+41​)dx, where [∙][\bullet][∙] denotes the greatest integer function, is
  1. (A)160(21π−1)\frac{1}{60}(21\pi - 1)601​(21π−1)
  2. (B)160(π−7)\frac{1}{60}(\pi - 7)601​(π−7)
  3. (C)760(3π−1)\frac{7}{60}(3\pi - 1)607​(3π−1)
  4. (D)760(π−3)\frac{7}{60}(\pi - 3)607​(π−3)

Correct answer: (C)

Step-by-step solution →
Q25·MathematicsNumericalJEE Main 2026
6∫0π∣(sin⁡3x+sin⁡2x+sin⁡x)∣dx6\int_{0}^{\pi}\left|(\sin 3x+\sin 2x+\sin x)\right|dx6∫0π​∣(sin3x+sin2x+sinx)∣dx is equal to……..

Correct answer: 17

Step-by-step solution →
Q26·MathematicsNumericalJEE Main 2026
If ∫014cot⁡−1(1−2x+4x2)dx=atan⁡−1(2)−blog⁡e(5)\int_{0}^{1} 4\cot^{-1}(1 - 2x + 4x^{2})dx = a\tan^{-1}(2) - b\log_{e}(5)∫01​4cot−1(1−2x+4x2)dx=atan−1(2)−bloge​(5), where a,b∈Na, b \in Na,b∈N, then (2a+b)(2a + b)(2a+b) is equal to _______.

Correct answer: 9

Step-by-step solution →
Q27·MathematicsSingle correctJEE Main 2026
The value of ∫−π/6π/6(π+4x111−sin⁡(∣x∣+π/6))dx\int_{-\pi/6}^{\pi/6}\left( \frac{\pi + 4x^{11}}{1 - \sin(|x| + \pi/6)} \right) dx∫−π/6π/6​(1−sin(∣x∣+π/6)π+4x11​)dx is equal to
  1. (A)2π2\pi2π
  2. (B)4π4\pi4π
  3. (C)8π8\pi8π
  4. (D)6π6\pi6π

Correct answer: (B)

Step-by-step solution →
Q28·MathematicsNumericalJEE Advanced 2025
If α=∫1/22tan⁡−1x2x2−3x+2 dx\alpha = \int_{1/2}^{2} \frac{\tan^{-1}x}{2x^2 - 3x + 2}\,dxα=∫1/22​2x2−3x+2tan−1x​dx, then the value of 7 tan⁡(2α7π)\sqrt{7}\,\tan\left( \frac{2\alpha\sqrt{7}}{\pi} \right)7​tan(π2α7​​) is ______ (Here, the inverse trigonometric function tan⁡−1x\tan^{-1}xtan−1x assumes values in (−π2,π2)\left( -\frac{\pi}{2}, \frac{\pi}{2} \right)(−2π​,2π​).)

Correct answer: 21

Step-by-step solution →
Q29·MathematicsSingle correctJEE Main 2025
The integral ∫−13/2( ∣π2xsin⁡(πx)∣ ) dx\displaystyle\int_{-1}^{3/2}\big(\,|\pi^2 x\sin(\pi x)|\,\big)\,dx∫−13/2​(∣π2xsin(πx)∣)dx is equal to:
  1. (A)3+2π3+2\pi3+2π
  2. (B)4+π4+\pi4+π
  3. (C)1+3π1+3\pi1+3π
  4. (D)2+3π2+3\pi2+3π

Correct answer: (C)

Step-by-step solution →
Q30·MathematicsSingle correctJEE Main 2025
Let f(x)f(x)f(x) be a positive function and I1=∫−1/212x f(2x(1−2x)) dxI_1=\displaystyle\int_{-1/2}^{1}2x\,f\big(2x(1-2x)\big)\,dxI1​=∫−1/21​2xf(2x(1−2x))dx and I2=∫−12f(x(1−x)) dxI_2=\displaystyle\int_{-1}^{2}f\big(x(1-x)\big)\,dxI2​=∫−12​f(x(1−x))dx. Then the value of I2I1\dfrac{I_2}{I_1}I1​I2​​ is equal to ___
  1. (A)999
  2. (B)666
  3. (C)121212
  4. (D)444

Correct answer: (D)

Step-by-step solution →
Q31·MathematicsSingle correctJEE Main 2025
The integral ∫0π(x+3)sin⁡x1+3cos⁡2x dx\displaystyle\int_0^\pi\dfrac{(x+3)\sin x}{1+3\cos^2x}\,dx∫0π​1+3cos2x(x+3)sinx​dx is equal to:
  1. (A)π3(π+1)\dfrac{\pi}{\sqrt{3}}(\pi+1)3​π​(π+1)
  2. (B)π3(π+2)\dfrac{\pi}{\sqrt{3}}(\pi+2)3​π​(π+2)
  3. (C)π33(π+6)\dfrac{\pi}{3\sqrt{3}}(\pi+6)33​π​(π+6)
  4. (D)π23(π+4)\dfrac{\pi}{2\sqrt{3}}(\pi+4)23​π​(π+4)

Correct answer: (C)

Step-by-step solution →
Q32·MathematicsSingle correctJEE Main 2025
Let f(x)+2f(1x)=x2+5f(x)+2f\left(\dfrac{1}{x}\right)=x^2+5f(x)+2f(x1​)=x2+5 and 2g(x)−3g(12)=x2g(x)-3g\left(\dfrac{1}{2}\right)=x2g(x)−3g(21​)=x, x>0x>0x>0. If α=∫12f(x) dx\alpha=\displaystyle\int_1^2 f(x)\,dxα=∫12​f(x)dx, and β=∫12g(x) dx\beta=\displaystyle\int_1^2 g(x)\,dxβ=∫12​g(x)dx, then the value of 9α+β9\alpha+\beta9α+β is:
  1. (A)1
  2. (B)0
  3. (C)10
  4. (D)11

Correct answer: (D)

Step-by-step solution →
Q33·MathematicsSingle correctJEE Main 2025
The value of ∫−11(1+∣x∣−x)ex+(∣x∣−x)e−xex+e−x dx\displaystyle\int_{-1}^{1}\dfrac{\left(1+\sqrt{|x|-x}\right)e^x+\left(\sqrt{|x|-x}\right)e^{-x}}{e^x+e^{-x}}\,dx∫−11​ex+e−x(1+∣x∣−x​)ex+(∣x∣−x​)e−x​dx is equal to:
  1. (A)3−2233-\dfrac{2\sqrt{2}}{3}3−322​​
  2. (B)2+2232+\dfrac{2\sqrt{2}}{3}2+322​​
  3. (C)1−2231-\dfrac{2\sqrt{2}}{3}1−322​​
  4. (D)1+2231+\dfrac{2\sqrt{2}}{3}1+322​​

Correct answer: (D)

Step-by-step solution →
Q34·MathematicsSingle correctJEE Main 2025
Let the domain of the function f(x)=log⁡2log⁡4log⁡6(3+4x−x2)f(x)=\log_2\log_4\log_6(3+4x-x^2)f(x)=log2​log4​log6​(3+4x−x2) be (a,b)(a,b)(a,b). If ∫0 b−a[x2] dx=p−q−r\displaystyle\int_0^{\,b-a}[x^2]\,dx=p-\sqrt q-\sqrt r∫0b−a​[x2]dx=p−q​−r​, p,q,r∈Np,q,r\in\mathbb{N}p,q,r∈N, gcd⁡(p,q,r)=1\gcd(p,q,r)=1gcd(p,q,r)=1, where [⋅][\cdot][⋅] is the greatest integer function, then p+q+rp+q+rp+q+r is equal to:
  1. (A)10
  2. (B)8
  3. (C)11
  4. (D)9

Correct answer: (A)

Step-by-step solution →
Q35·MathematicsSingle correctJEE Main 2025
The integral ∫0π/28x dx4cos⁡2x+sin⁡2x\displaystyle\int_0^{\pi/2}\dfrac{8x\,dx}{4\cos^2 x+\sin^2 x}∫0π/2​4cos2x+sin2x8xdx​ is equal to:
  1. (A)2π22\pi^22π2
  2. (B)4π24\pi^24π2
  3. (C)π2\pi^2π2
  4. (D)3π22\dfrac{3\pi^2}{2}23π2​

Correct answer: (A)

Step-by-step solution →
Q36·MathematicsSingle correctJEE Main 2025
Let f:[1,∞)→[2,∞)f:[1,\infty)\to[2,\infty)f:[1,∞)→[2,∞) be a differentiable function. If 10∫1xf(t) dt=5x f(x)−x5−910\displaystyle\int_1^x f(t)\,dt = 5x\,f(x) - x^5 - 910∫1x​f(t)dt=5xf(x)−x5−9 for all x≥1x\ge 1x≥1, then the value of f(3)f(3)f(3) is:
  1. (A)18
  2. (B)32
  3. (C)22
  4. (D)26

Correct answer: (B)

Step-by-step solution →
Q37·MathematicsIntegerJEE Main 2025
Let [⋅][\cdot][⋅] denote the greatest integer function. If ∫0e2[1ex−1]dx=α−log⁡e2\displaystyle\int_0^{e^2}\left[\dfrac{1}{e^{x-1}}\right]dx=\alpha-\log_e 2∫0e2​[ex−11​]dx=α−loge​2, then α3\alpha^3α3 is equal to ______.

Correct answer: 8

Step-by-step solution →
Q38·MathematicsSingle correctJEE Main 2025
4∫01(13+x2+1+x2)dx−3log⁡e(3)4\displaystyle\int_0^1\left(\dfrac{1}{\sqrt{3+x^2}+\sqrt{1+x^2}}\right)dx-3\log_e(\sqrt3)4∫01​(3+x2​+1+x2​1​)dx−3loge​(3​) is equal to:
  1. (A)2+2+log⁡e(1+2)2+\sqrt2+\log_e(1+\sqrt2)2+2​+loge​(1+2​)
  2. (B)2−2−log⁡e(1+2)2-\sqrt2-\log_e(1+\sqrt2)2−2​−loge​(1+2​)
  3. (C)2+2−log⁡e(1+2)2+\sqrt2-\log_e(1+\sqrt2)2+2​−loge​(1+2​)
  4. (D)2−2+log⁡e(1+2)2-\sqrt2+\log_e(1+\sqrt2)2−2​+loge​(1+2​)

Correct answer: (B)

Step-by-step solution →
Q39·MathematicsSingle correctJEE Main 2025
Let (a,b)(a,b)(a,b) be the point of intersection of the curve x2=2yx^2=2yx2=2y and the straight line y−2x−6=0y-2x-6=0y−2x−6=0 in the second quadrant. Then the integral I=∫ab9x21+5x dxI=\displaystyle\int_a^b \dfrac{9x^2}{1+5^x}\,dxI=∫ab​1+5x9x2​dx is equal to:
  1. (A)24
  2. (B)27
  3. (C)18
  4. (D)21

Correct answer: (A)

Step-by-step solution →
Q40·MathematicsSingle correctJEE Main 2025
Let f′(x)=∫0xt(t2−9t+20) dtf'(x)=\displaystyle\int_0^x t(t^2-9t+20)\,dtf′(x)=∫0x​t(t2−9t+20)dt, 1≤x≤51\le x\le 51≤x≤5. If the range of f is [α,β][\alpha,\beta][α,β], then 4(α+β)4(\alpha+\beta)4(α+β) equals:
  1. (A)157
  2. (B)253
  3. (C)125
  4. (D)154

Correct answer: (A)

Step-by-step solution →
Q41·MathematicsSingle correctJEE Main 2025
The integral 80∫0π/2(sin⁡θ+cos⁡θ9+16sin⁡2θ)dθ80\displaystyle\int_0^{\pi/2}\left(\dfrac{\sin\theta+\cos\theta}{9+16\sin 2\theta}\right)d\theta80∫0π/2​(9+16sin2θsinθ+cosθ​)dθ is equal to:
  1. (A)3log⁡e43\log_e 43loge​4
  2. (B)6log⁡e46\log_e 46loge​4
  3. (C)4log⁡e34\log_e 34loge​3
  4. (D)2log⁡e32\log_e 32loge​3

Correct answer: (C)

Step-by-step solution →
Q42·MathematicsIntegerJEE Main 2025
Let f:(0,∞)→Rf:(0,\infty)\to Rf:(0,∞)→R be a twice differentiable function. If for some a≠0a\ne 0a=0, ∫01f(λx) dλ=a f(x)\displaystyle\int_0^1 f(\lambda x)\,d\lambda=a\,f(x)∫01​f(λx)dλ=af(x), f(1)=1f(1)=1f(1)=1 and f(16)=18f(16)=\dfrac{1}{8}f(16)=81​, then 16−f′(116)16-f'\left(\dfrac{1}{16}\right)16−f′(161​) is equal to ______.

Correct answer: 112

Step-by-step solution →
Q43·MathematicsIntegerJEE Main 2025
If 24∫0π/4(sin⁡∣4x−π12∣+[2sin⁡x])dx=2π+α24\displaystyle\int_0^{\pi/4}\left(\sin\left|4x-\dfrac{\pi}{12}\right|+[2\sin x]\right)dx=2\pi+\alpha24∫0π/4​(sin​4x−12π​​+[2sinx])dx=2π+α, where [⋅][\cdot][⋅] denotes the greatest integer function, then α\alphaα is equal to ______.

Correct answer: 12

Step-by-step solution →
Q44·MathematicsSingle correctJEE Main 2025
Let f:R→Rf:R\to Rf:R→R be a twice differentiable function such that f(2)=1f(2)=1f(2)=1. If F(x)=x f(x)F(x)=x\,f(x)F(x)=xf(x) for all x∈Rx\in Rx∈R, ∫02x F′(x)dx=6\int_0^2 x\,F'(x)dx=6∫02​xF′(x)dx=6 and ∫02x2F′′(x)dx=40\int_0^2 x^2 F''(x)dx=40∫02​x2F′′(x)dx=40, then F′(2)+∫02F(x)dxF'(2)+\int_0^2 F(x)dxF′(2)+∫02​F(x)dx is equal to:
  1. (A)11
  2. (B)15
  3. (C)9
  4. (D)13

Correct answer: (A)

Step-by-step solution →
Q45·MathematicsSingle correctJEE Main 2025
If ∫−π/2π/296x2cos⁡2x1+exdx=π(απ2+β)\int_{-\pi/2}^{\pi/2}\frac{96x^2\cos^2 x}{1+e^x}dx=\pi(\alpha\pi^2+\beta)∫−π/2π/2​1+ex96x2cos2x​dx=π(απ2+β), α,β∈Z\alpha, \beta\in Zα,β∈Z, then (α+β)2(\alpha+\beta)^2(α+β)2 equals:
  1. (A)144
  2. (B)196
  3. (C)100
  4. (D)64

Correct answer: (C)

Step-by-step solution →
Q46·MathematicsSingle correctJEE Main 2025
Let f be a real valued continuous function defined on the positive real axis such that g(x)=∫0xt f(t)dtg(x)=\int_0^x t\,f(t)dtg(x)=∫0x​tf(t)dt. If g(x3)=x6+x7g(x^3)=x^6+x^7g(x3)=x6+x7, then value of ∑r=115f(r3)\sum_{r=1}^{15} f(r^3)∑r=115​f(r3) is:
  1. (A)320
  2. (B)340
  3. (C)270
  4. (D)310

Correct answer: (D)

Step-by-step solution →
Q47·MathematicsSingle correctJEE Main 2025
In I(m,n)=∫01xm−1(1−x)n−1dxI(m,n)=\int_0^1 x^{m-1}(1-x)^{n-1}dxI(m,n)=∫01​xm−1(1−x)n−1dx, m,n>0m,n>0m,n>0, then I(9,14)+I(10,13)I(9,14)+I(10,13)I(9,14)+I(10,13) is
  1. (A)I(9,1)I(9,1)I(9,1)
  2. (B)I(19,27)I(19,27)I(19,27)
  3. (C)I(1,13)I(1,13)I(1,13)
  4. (D)I(9,13)I(9,13)I(9,13)

Correct answer: (D)

Step-by-step solution →
Q48·MathematicsSingle correctJEE Main 2025
The value of ∫e2e41x(e((log⁡ex)2+1)−1e((log⁡ex)2+1)−1+e((6−log⁡ex)2+1)−1)dx\displaystyle\int_{e^2}^{e^4}\frac{1}{x}\left(\frac{e^{\left((\log_e x)^2+1\right)^{-1}}}{e^{\left((\log_e x)^2+1\right)^{-1}}+e^{\left((6-\log_e x)^2+1\right)^{-1}}}\right)dx∫e2e4​x1​(e((loge​x)2+1)−1+e((6−loge​x)2+1)−1e((loge​x)2+1)−1​)dx is
  1. (A)log⁡e2\log_e 2loge​2
  2. (B)222
  3. (C)111
  4. (D)e2e^2e2

Correct answer: (C)

Step-by-step solution →
Q49·MathematicsSingle correctJEE Main 2025
If I=∫0π/2sin⁡2xsin⁡4x+cos⁡4x dxI=\displaystyle\int_0^{\pi/2}\dfrac{\sin^2 x}{\sin^4 x+\cos^4 x}\,dxI=∫0π/2​sin4x+cos4xsin2x​dx, then ∫0π/2xsin⁡xcos⁡xsin⁡4x+cos⁡4x dx\displaystyle\int_0^{\pi/2}\dfrac{x\sin x\cos x}{\sin^4 x+\cos^4 x}\,dx∫0π/2​sin4x+cos4xxsinxcosx​dx equals :
  1. (A)π216\dfrac{\pi^2}{16}16π2​
  2. (B)π24\dfrac{\pi^2}{4}4π2​
  3. (C)π28\dfrac{\pi^2}{8}8π2​
  4. (D)π212\dfrac{\pi^2}{12}12π2​

Correct answer: (A)

Step-by-step solution →
Q50·MathematicsSingle correctJEE Main 2025
Let f(x)=7tan⁡8x+7tan⁡6x−3tan⁡4x−3tan⁡2xf(x)=7\tan^8x+7\tan^6x-3\tan^4x-3\tan^2xf(x)=7tan8x+7tan6x−3tan4x−3tan2x, I1=∫0π/4f(x) dxI_1=\int_0^{\pi/4}f(x)\,dxI1​=∫0π/4​f(x)dx, I2=∫0π/4x f(x) dxI_2=\int_0^{\pi/4}x\,f(x)\,dxI2​=∫0π/4​xf(x)dx. Then 7I1+12I27I_1+12I_27I1​+12I2​ is:
  1. (A)2π2\pi2π
  2. (B)π\piπ
  3. (C)1
  4. (D)2

Correct answer: (C)

Step-by-step solution →
Q51·MathematicsIntegerJEE Advanced 2024
Let the function f:[1,∞)→Rf : [1, \infty) \to \mathbb{R}f:[1,∞)→R be defined by f(t)={(−1)n+12; if t=2n−1,n∈N(2n+1−t)2f(2n−1)+(t−(2n−1))2f(2n+1); if 2n−1<t<2n+1,n∈Nf(t) = \begin{cases} (-1)^{n+1} 2 & ; \text{ if } t = 2n - 1, n \in \mathbb{N} \\ \frac{(2n + 1 - t)}{2} f(2n - 1) + \frac{(t - (2n - 1))}{2} f(2n + 1) & ; \text{ if } 2n - 1 < t < 2n + 1, n \in \mathbb{N} \end{cases}f(t)={(−1)n+122(2n+1−t)​f(2n−1)+2(t−(2n−1))​f(2n+1)​; if t=2n−1,n∈N; if 2n−1<t<2n+1,n∈N​. Define g(x)=∫1xf(t) dtg(x) = \int_1^x f(t) \, dtg(x)=∫1x​f(t)dt, x∈(1,∞)x \in (1, \infty)x∈(1,∞). Let α\alphaα denote the number of solutions of the equation g(x) = 0 in the interval (1,8](1, 8](1,8] and β=lim⁡x→1+g(x)x−1\beta = \lim_{x \to 1^+} \frac{g(x)}{x - 1}β=limx→1+​x−1g(x)​. Then the value of α+β\alpha + \betaα+β is equal to ______

Correct answer: 5

Step-by-step solution →
Q52·MathematicsNumericalJEE Advanced 2024
Let f:[0,π2]→[0,1]f : \left[0, \frac{\pi}{2}\right] \to [0, 1]f:[0,2π​]→[0,1] be the function defined by f(x)=sin⁡2xf(x) = \sin^2 xf(x)=sin2x and let g:[0,π2]→[0,∞)g : \left[0, \frac{\pi}{2}\right] \to [0, \infty)g:[0,2π​]→[0,∞) be the function defined by g=πx2−x2g = \sqrt{\frac{\pi x}{2} - x^2}g=2πx​−x2​. The value of 16π3∫0π2f(x)g(x) dx\frac{16}{\pi^3} \int_0^{\frac{\pi}{2}} f(x)g(x) \, dxπ316​∫02π​​f(x)g(x)dx is ______

Correct answer: 0.25

Step-by-step solution →
Q53·MathematicsNumericalJEE Advanced 2024
Let f:[0,π2]→[0,1]f : \left[0, \frac{\pi}{2}\right] \to [0, 1]f:[0,2π​]→[0,1] be the function defined by f(x)=sin⁡2xf(x) = \sin^2 xf(x)=sin2x and let g:[0,π2]→[0,∞)g : \left[0, \frac{\pi}{2}\right] \to [0, \infty)g:[0,2π​]→[0,∞) be the function defined by g=πx2−x2g = \sqrt{\frac{\pi x}{2} - x^2}g=2πx​−x2​. The value of 2∫0π2f(x)g(x) dx−∫0π2g(x) dx2\int_0^{\frac{\pi}{2}} f(x)g(x) \, dx - \int_0^{\frac{\pi}{2}} g(x) \, dx2∫02π​​f(x)g(x)dx−∫02π​​g(x)dx is ______

Correct answer: 0.00

Step-by-step solution →
Q54·MathematicsNumericalJEE Main 2024
Let lim⁡n→∞(nn4+1−2n(n2+1)n4+1+nn4+16−8n(n2+4)n4+16+⋯+nn4+n4−2n⋅n2(n2+n2)n4+n4)\lim_{n \to \infty} \left( \frac{n}{\sqrt{n^4+1}} - \frac{2n}{(n^2+1)\sqrt{n^4+1}} + \frac{n}{\sqrt{n^4+16}} - \frac{8n}{(n^2+4)\sqrt{n^4+16}} + \cdots + \frac{n}{\sqrt{n^4+n^4}} - \frac{2n \cdot n^2}{(n^2+n^2)\sqrt{n^4+n^4}} \right)limn→∞​(n4+1​n​−(n2+1)n4+1​2n​+n4+16​n​−(n2+4)n4+16​8n​+⋯+n4+n4​n​−(n2+n2)n4+n4​2n⋅n2​) be πk\frac{\pi}{k}kπ​, using only the principal values of the inverse trigonometric functions. Then k2k^2k2 is equal to ________.

Correct answer: 32

Step-by-step solution →
Q55·MathematicsSingle correctJEE Main 2024
The integral ∫1/43/4cos⁡(2cot⁡−11−x1+x)dx\displaystyle\int_{1/4}^{3/4}\cos\left(2\cot^{-1}\sqrt{\dfrac{1-x}{1+x}}\right)dx∫1/43/4​cos(2cot−11+x1−x​​)dx is equal to:
  1. (A)−12-\dfrac{1}{2}−21​
  2. (B)14\dfrac{1}{4}41​
  3. (C)12\dfrac{1}{2}21​
  4. (D)−14-\dfrac{1}{4}−41​

Correct answer: (D)

Step-by-step solution →
Q56·MathematicsSingle correctJEE Main 2024
The value of the integral ∫−12log⁡e(x+x2+1)dx\displaystyle\int_{-1}^{2}\log_{e}\left(x+\sqrt{x^{2}+1}\right)dx∫−12​loge​(x+x2+1​)dx is:
  1. (A)5−2+log⁡e(9+451+2)\sqrt{5}-\sqrt{2}+\log_{e}\left(\dfrac{9+4\sqrt{5}}{1+\sqrt{2}}\right)5​−2​+loge​(1+2​9+45​​)
  2. (B)2−5+log⁡e(9+451+2)\sqrt{2}-\sqrt{5}+\log_{e}\left(\dfrac{9+4\sqrt{5}}{1+\sqrt{2}}\right)2​−5​+loge​(1+2​9+45​​)
  3. (C)5−2+log⁡e(7+451+2)\sqrt{5}-\sqrt{2}+\log_{e}\left(\dfrac{7+4\sqrt{5}}{1+\sqrt{2}}\right)5​−2​+loge​(1+2​7+45​​)
  4. (D)2−5+log⁡e(7+451+2)\sqrt{2}-\sqrt{5}+\log_{e}\left(\dfrac{7+4\sqrt{5}}{1+\sqrt{2}}\right)2​−5​+loge​(1+2​7+45​​)

Correct answer: (B)

Step-by-step solution →
Q57·MathematicsSingle correctJEE Main 2024
Let ∫alog⁡e4dxex−1=π6\displaystyle\int_{a}^{\log_e 4}\dfrac{dx}{\sqrt{e^x-1}}=\dfrac{\pi}{6}∫aloge​4​ex−1​dx​=6π​. Then eae^aea and e−ae^{-a}e−a are the roots of the equation
  1. (A)2x2−5x+2=02x^2-5x+2=02x2−5x+2=0
  2. (B)x2−2x−8=0x^2-2x-8=0x2−2x−8=0
  3. (C)2x2−5x−2=02x^2-5x-2=02x2−5x−2=0
  4. (D)x2+2x−8=0x^2+2x-8=0x2+2x−8=0

Correct answer: (A)

Step-by-step solution →
Q58·MathematicsSingle correctJEE Main 2024
The value of k∈Nk\in\mathbb{N}k∈N for which the integral In=∫01(1−xk)n dxI_n=\displaystyle\int_0^1(1-x^k)^n\,dxIn​=∫01​(1−xk)ndx, n∈Nn\in\mathbb{N}n∈N, satisfies 147 I20=148 I21147\,I_{20}=148\,I_{21}147I20​=148I21​ is:
  1. (A)101010
  2. (B)888
  3. (C)141414
  4. (D)777

Correct answer: (D)

Step-by-step solution →
Q59·MathematicsSingle correctJEE Main 2024
∫0π/4cos⁡2x sin⁡2x(cos⁡3x+sin⁡3x)2 dx\displaystyle\int_{0}^{\pi/4}\frac{\cos^{2}x\,\sin^{2}x}{(\cos^{3}x+\sin^{3}x)^{2}}\,dx∫0π/4​(cos3x+sin3x)2cos2xsin2x​dx is equal to
  1. (A)112\tfrac{1}{12}121​
  2. (B)13\tfrac{1}{3}31​
  3. (C)16\tfrac{1}{6}61​
  4. (D)19\tfrac{1}{9}91​

Correct answer: (C)

Step-by-step solution →
Q60·MathematicsNumericalJEE Main 2024
Let rk=∫01(1−x7)k dx∫01(1−x7)k+1 dxr_{k}=\dfrac{\int_{0}^{1}(1-x^{7})^{k}\,dx}{\int_{0}^{1}(1-x^{7})^{k+1}\,dx}rk​=∫01​(1−x7)k+1dx∫01​(1−x7)kdx​, k∈Nk\in\mathbb{N}k∈N. Then the value of ∑k=11017(rk−1)\displaystyle\sum_{k=1}^{10}\frac{1}{7(r_{k}-1)}k=1∑10​7(rk​−1)1​ is equal to _______.

Correct answer: 65

Step-by-step solution →
Q61·MathematicsNumericalJEE Main 2024
Let [t][t][t] denote the greatest integer less than or equal to ttt. Let f:[0,∞)→Rf:[0,\infty)\to Rf:[0,∞)→R be a function defined by f(x)=[x2+3]−[x]f(x)=\left[\frac{x}{2}+3\right]-\left[\sqrt{x}\right]f(x)=[2x​+3]−[x​]. Let SSS be the set of all points in the interval [0,8][0,8][0,8] at which fff is not continuous. Then ∑a∈Sa\sum_{a\in S}a∑a∈S​a is equal to ______.

Correct answer: 17

Step-by-step solution →
Q62·MathematicsNumericalJEE Main 2024
Let [t][t][t] denote the largest integer less than or equal to ttt. If ∫03([x2]+[x22])dx=a+b2−3−5+c6−7\int_{0}^{3}\left([x^{2}]+\left[\frac{x^{2}}{2}\right]\right)dx=a+b\sqrt{2}-\sqrt{3}-\sqrt{5}+c\sqrt{6}-\sqrt{7}∫03​([x2]+[2x2​])dx=a+b2​−3​−5​+c6​−7​, where a,b,c∈Za,b,c\in Za,b,c∈Z, then a+b+ca+b+ca+b+c is equal to ______.

Correct answer: 23

Step-by-step solution →
Q63·MathematicsNumericalJEE Main 2024
If f(t)=∫0π2x dx1−cos⁡2t sin⁡2xf(t)=\displaystyle\int_0^{\pi}\dfrac{2x\,dx}{1-\cos^2 t\,\sin^2 x}f(t)=∫0π​1−cos2tsin2x2xdx​, 0<t<π0<t<\pi0<t<π, then the value of ∫0π/2π2 dtf(t)\displaystyle\int_0^{\pi/2}\dfrac{\pi^2\,dt}{f(t)}∫0π/2​f(t)π2dt​ equals __________.

Correct answer: 1

Step-by-step solution →
Q64·MathematicsSingle correctJEE Main 2024
The integral ∫0π/4136sin⁡x3sin⁡x+5cos⁡x dx\displaystyle\int_{0}^{\pi/4} \dfrac{136\sin x}{3\sin x + 5\cos x}\,dx∫0π/4​3sinx+5cosx136sinx​dx is equal to:
  1. (A)3π−50log⁡e2+20log⁡e53\pi - 50\log_e 2 + 20\log_e 53π−50loge​2+20loge​5
  2. (B)3π−25log⁡e2+10log⁡e53\pi - 25\log_e 2 + 10\log_e 53π−25loge​2+10loge​5
  3. (C)3π−10log⁡e(22)+10log⁡e53\pi - 10\log_e\left(2\sqrt{2}\right) + 10\log_e 53π−10loge​(22​)+10loge​5
  4. (D)3π−30log⁡e2+20log⁡e53\pi - 30\log_e 2 + 20\log_e 53π−30loge​2+20loge​5

Correct answer: (A)

Step-by-step solution →
Q65·MathematicsSingle correctJEE Main 2024
Let β(m,n)=∫01xm−1(1−x)n−1 dx\beta(m,n)=\displaystyle\int_0^1 x^{m-1}(1-x)^{n-1}\,dxβ(m,n)=∫01​xm−1(1−x)n−1dx, m,n>0m,n>0m,n>0. If ∫01(1−x10)20 dx=a β(b,c)\displaystyle\int_0^1 (1-x^{10})^{20}\,dx=a\,\beta(b,c)∫01​(1−x10)20dx=aβ(b,c), then 100(a+b+c)100(a+b+c)100(a+b+c) equals:
  1. (A)1021
  2. (B)1120
  3. (C)2012
  4. (D)2120

Correct answer: (D)

Step-by-step solution →
Q66·MathematicsSingle correctJEE Main 2024
The value of ∫−ππ2y(1+sin⁡y)1+cos⁡2y dy\displaystyle\int_{-\pi}^{\pi} \dfrac{2y(1+\sin y)}{1+\cos^2 y}\,dy∫−ππ​1+cos2y2y(1+siny)​dy is:
  1. (A)π2\pi^2π2
  2. (B)π22\dfrac{\pi^2}{2}2π2​
  3. (C)π2\dfrac{\pi}{2}2π​
  4. (D)2π22\pi^22π2

Correct answer: (A)

Step-by-step solution →
Q67·MathematicsSingle correctJEE Main 2024
Let f(x)={−2, −2≤x≤0x−2, 0<x≤2f(x)=\begin{cases}-2 & ,\ -2\le x\le0\\ x-2 & ,\ 0<x\le2\end{cases}f(x)={−2x−2​, −2≤x≤0, 0<x≤2​ and h(x)=f(∣x∣)+∣f(x)∣h(x)=f(|x|)+|f(x)|h(x)=f(∣x∣)+∣f(x)∣. Then ∫−22h(x) dx\displaystyle\int_{-2}^{2}h(x)\,dx∫−22​h(x)dx is equal to:
  1. (A)222
  2. (B)444
  3. (C)111
  4. (D)666

Correct answer: (A)

Step-by-step solution →
Q68·MathematicsNumericalJEE Main 2024
If ∫0π/4sin⁡2x1+sin⁡xcos⁡x dx=1alog⁡e(a3)+πb3\displaystyle\int_0^{\pi/4}\dfrac{\sin^2 x}{1+\sin x\cos x}\,dx=\dfrac1a\log_e\left(\dfrac a3\right)+\dfrac{\pi}{b\sqrt3}∫0π/4​1+sinxcosxsin2x​dx=a1​loge​(3a​)+b3​π​, where a,b∈Na,b\in\mathbb{N}a,b∈N, then a+ba+ba+b is equal to ___

Correct answer: 8

Step-by-step solution →
Q69·MathematicsSingle correctJEE Main 2024
If the value of the integral ∫−11cos⁡αx1+3x dx\displaystyle\int_{-1}^{1}\dfrac{\cos\alpha x}{1+3^x}\,dx∫−11​1+3xcosαx​dx is 2π\dfrac{2}{\pi}π2​, then a value of α\alphaα is
  1. (A)π6\tfrac{\pi}{6}6π​
  2. (B)π2\tfrac{\pi}{2}2π​
  3. (C)π3\tfrac{\pi}{3}3π​
  4. (D)π4\tfrac{\pi}{4}4π​

Correct answer: (B)

Step-by-step solution →
Q70·MathematicsSingle correctJEE Main 2024
The value of the integral ∫0π/4x dxsin⁡4(2x)+cos⁡4(2x)\displaystyle\int_0^{\pi/4}\dfrac{x\,dx}{\sin^4(2x)+\cos^4(2x)}∫0π/4​sin4(2x)+cos4(2x)xdx​ equals:
  1. (A)2 π28\dfrac{\sqrt2\,\pi^2}{8}82​π2​
  2. (B)2 π216\dfrac{\sqrt2\,\pi^2}{16}162​π2​
  3. (C)2 π232\dfrac{\sqrt2\,\pi^2}{32}322​π2​
  4. (D)2 π264\dfrac{\sqrt2\,\pi^2}{64}642​π2​

Correct answer: (C)

Step-by-step solution →
Q71·MathematicsNumericalJEE Main 2024
Let f:(0,∞)→Rf:(0,\infty)\to\mathbb Rf:(0,∞)→R and F(x)=∫0xtf(t) dtF(x)=\displaystyle\int_0^x t f(t)\,dtF(x)=∫0x​tf(t)dt. If F(x2)=x4+x5F(x^2)=x^4+x^5F(x2)=x4+x5, then ∑r=112f(r2)\displaystyle\sum_{r=1}^{12} f(r^2)r=1∑12​f(r2) is equal to __________.

Correct answer: 219

Step-by-step solution →
Q72·MathematicsSingle correctJEE Main 2024
If ∫0π/3cos⁡4x dx=aπ+b3\displaystyle\int_0^{\pi/3}\cos^4 x\,dx=a\pi+b\sqrt 3∫0π/3​cos4xdx=aπ+b3​, where aaa and bbb are rational numbers, then 9a+8b9a+8b9a+8b is equal to:
  1. (A)2
  2. (B)1
  3. (C)3
  4. (D)32\dfrac{3}{2}23​

Correct answer: (A)

Step-by-step solution →
Q73·MathematicsNumericalJEE Main 2024
If ∫−π/2π/282cos⁡x dx(1+esin⁡x)(1+sin⁡4x)=απ+βlog⁡e(3+22)\displaystyle\int_{-\pi/2}^{\pi/2}\dfrac{8\sqrt2\cos x\,dx}{(1+e^{\sin x})(1+\sin^4 x)}=\alpha\pi+\beta\log_e(3+2\sqrt2)∫−π/2π/2​(1+esinx)(1+sin4x)82​cosxdx​=απ+βloge​(3+22​), where α,β\alpha,\betaα,β are integers, then α2+β2\alpha^2+\beta^2α2+β2 equals ___

Correct answer: 8

Step-by-step solution →
Q74·MathematicsSingle correctJEE Main 2024
The value of ∫01(2x3−3x2−x+1)1/3 dx\displaystyle\int_0^1 (2x^3-3x^2-x+1)^{1/3}\,dx∫01​(2x3−3x2−x+1)1/3dx is equal to:
  1. (A)0
  2. (B)1
  3. (C)2
  4. (D)−1-1−1

Correct answer: (A)

Step-by-step solution →
Q75·MathematicsNumericalJEE Main 2024
∣120π3∫0πx2sin⁡xcos⁡xsin⁡4x+cos⁡4x dx∣\left|\dfrac{120}{\pi^3}\displaystyle\int_0^{\pi}\dfrac{x^2\sin x\cos x}{\sin^4 x+\cos^4 x}\,dx\right|​π3120​∫0π​sin4x+cos4xx2sinxcosx​dx​ is equal to ______.

Correct answer: 15

Step-by-step solution →
Q76·MathematicsNumericalJEE Main 2024
Let f:R→Rf:R\to Rf:R→R be a function defined by f(x)=4x4x+2f(x)=\dfrac{4^x}{4^x+2}f(x)=4x+24x​ and M=∫f(a)f(1−a)xsin⁡4(x(1−x))dxM=\displaystyle\int_{f(a)}^{f(1-a)} x\sin^4\left(x(1-x)\right)dxM=∫f(a)f(1−a)​xsin4(x(1−x))dx, N=∫f(a)f(1−a)sin⁡4(x(1−x))dxN=\displaystyle\int_{f(a)}^{f(1-a)}\sin^4\left(x(1-x)\right)dxN=∫f(a)f(1−a)​sin4(x(1−x))dx; a≠12a\ne\dfrac{1}{2}a=21​. If αM=βN\alpha M=\beta NαM=βN, α,β∈N\alpha,\beta\in Nα,β∈N, then the least value of α2+β2\alpha^2+\beta^2α2+β2 is equal to ______.

Correct answer: 5

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Q77·MathematicsNumericalJEE Main 2024
If the integral 525∫0π/2sin⁡2xcos⁡11/2x(1+cos⁡5/2x)1/2dx525\displaystyle\int_0^{\pi/2}\sin 2x\cos^{11/2}x\left(1+\cos^{5/2}x\right)^{1/2}dx525∫0π/2​sin2xcos11/2x(1+cos5/2x)1/2dx is equal to (n2−64)\left(n\sqrt{2}-64\right)(n2​−64), then n is equal to ______.

Correct answer: 176

Step-by-step solution →
Q78·MathematicsSingle correctJEE Main 2024
The value of lim⁡n→∞∑k=1nn3(n2+k2)(n2+3k2)\displaystyle\lim_{n\to\infty}\sum_{k=1}^{n}\dfrac{n^3}{(n^2+k^2)(n^2+3k^2)}n→∞lim​k=1∑n​(n2+k2)(n2+3k2)n3​ is:
  1. (A)(23−3)π24\dfrac{(2\sqrt3-3)\pi}{24}24(23​−3)π​
  2. (B)13π8(43+3)\dfrac{13\pi}{8(4\sqrt3+3)}8(43​+3)13π​
  3. (C)13(23−3)π24\dfrac{13(2\sqrt3-3)\pi}{24}2413(23​−3)π​
  4. (D)π8(23+3)\dfrac{\pi}{8(2\sqrt3+3)}8(23​+3)π​

Correct answer: (B)

Step-by-step solution →
Q79·MathematicsSingle correctJEE Main 2024
Let y=f(x)y=f(x)y=f(x) be a thrice differentiable function in (−5,5)(-5,5)(−5,5). Let the tangents to the curve y=f(x)y=f(x)y=f(x) at (1,f(1))(1,f(1))(1,f(1)) and (3,f(3))(3,f(3))(3,f(3)) make angles π6\dfrac{\pi}{6}6π​ and π4\dfrac{\pi}{4}4π​ respectively with positive x-axis. If 27∫13((f′(t))2+1)f′′(t) dt=α+β327\displaystyle\int_1^3\left((f'(t))^2+1\right)f''(t)\,dt=\alpha+\beta\sqrt{3}27∫13​((f′(t))2+1)f′′(t)dt=α+β3​ where α,β\alpha,\betaα,β are integers, then the value of α+β\alpha+\betaα+β equals:
  1. (A)−14-14−14
  2. (B)262626
  3. (C)−16-16−16
  4. (D)363636

Correct answer: (B)

Step-by-step solution →
Q80·MathematicsNumericalJEE Main 2024
The value 9∫09[10xx+1 ]dx9\displaystyle\int_0^9\left[\sqrt{\dfrac{10x}{x+1}}\,\right]dx9∫09​[x+110x​​]dx, where [t][t][t] denotes the greatest integer less than or equal to ttt, is ___

Correct answer: 155

Step-by-step solution →
Q81·MathematicsSingle correctJEE Main 2024
Let f:R→Rf:R\to Rf:R→R be a function defined f(x)=x(1+x4)1/4f(x)=\dfrac{x}{(1+x^4)^{1/4}}f(x)=(1+x4)1/4x​ and g(x)=f(f(f(f(x))))g(x)=f(f(f(f(x))))g(x)=f(f(f(f(x)))) then 18∫025x2 g(x) dx18\displaystyle\int_0^{\sqrt{2\sqrt{5}}}x^2\,g(x)\,dx18∫025​​​x2g(x)dx
  1. (A)333333
  2. (B)363636
  3. (C)424242
  4. (D)393939

Correct answer: (D)

Step-by-step solution →
Q82·MathematicsSingle correctJEE Main 2024
Let f:R→Rf:R\to Rf:R→R be defined f(x)=ae2x+bex+cxf(x)=ae^{2x}+be^{x}+cxf(x)=ae2x+bex+cx. If f(0)=−1f(0)=-1f(0)=−1, f′(log⁡e2)=21f'(\log_e 2)=21f′(loge​2)=21 and ∫0log⁡e4(f(x)−cx) dx=392\displaystyle\int_0^{\log_e 4}(f(x)-cx)\,dx=\dfrac{39}{2}∫0loge​4​(f(x)−cx)dx=239​, then the value of ∣a+b+c∣|a+b+c|∣a+b+c∣ equals:
  1. (A)161616
  2. (B)101010
  3. (C)121212
  4. (D)888

Correct answer: (D)

Step-by-step solution →
Q83·MathematicsSingle correctJEE Main 2024
If the value of the integral ∫−π/2π/2(x2cos⁡x1+πx+1+sin⁡2x1+esin⁡x2023)dx=π4(π+a)−2\displaystyle\int_{-\pi/2}^{\pi/2}\left(\dfrac{x^2\cos x}{1+\pi^x}+\dfrac{1+\sin^2 x}{1+e^{\sin x^{2023}}}\right)dx=\dfrac{\pi}{4}(\pi+a)-2∫−π/2π/2​(1+πxx2cosx​+1+esinx20231+sin2x​)dx=4π​(π+a)−2, then the value of a is
  1. (A)333
  2. (B)−32-\dfrac{3}{2}−23​
  3. (C)222
  4. (D)32\dfrac{3}{2}23​

Correct answer: (A)

Step-by-step solution →
Q84·MathematicsNumericalJEE Main 2024
If ∫π/6π/31−sin⁡2x dx=α+β2+γ3\displaystyle\int_{\pi/6}^{\pi/3}\sqrt{1-\sin 2x}\,dx=\alpha+\beta\sqrt2+\gamma\sqrt3∫π/6π/3​1−sin2x​dx=α+β2​+γ3​, where α,β\alpha,\betaα,β and γ\gammaγ are rational numbers, then 3α+4β−γ3\alpha+4\beta-\gamma3α+4β−γ is equal to ___.

Correct answer: 6

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Q85·MathematicsSingle correctJEE Main 2024
For 0<a<10<a<10<a<1, the value of the integral ∫0πdx1−2acos⁡x+a2\displaystyle\int_{0}^{\pi}\dfrac{dx}{1-2a\cos x+a^2}∫0π​1−2acosx+a2dx​ is :
  1. (A)π2π+a2\dfrac{\pi^2}{\pi+a^2}π+a2π2​
  2. (B)π2π−a2\dfrac{\pi^2}{\pi-a^2}π−a2π2​
  3. (C)π1−a2\dfrac{\pi}{1-a^2}1−a2π​
  4. (D)π1+a2\dfrac{\pi}{1+a^2}1+a2π​

Correct answer: (C)

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Q86·MathematicsSingle correctJEE Main 2024
If (a,b)(a,b)(a,b) be the orthocentre of the triangle whose vertices are (1,2)(1,2)(1,2), (2,3)(2,3)(2,3) and (3,1)(3,1)(3,1) and I1=∫abxsin⁡(4x−x2) dxI_1=\displaystyle\int_a^b x\sin(4x-x^2)\,dxI1​=∫ab​xsin(4x−x2)dx, I2=∫absin⁡(4x−x2) dxI_2=\displaystyle\int_a^b \sin(4x-x^2)\,dxI2​=∫ab​sin(4x−x2)dx, then 36 I1I236\,\dfrac{I_1}{I_2}36I2​I1​​ is equal to:
  1. (A)72
  2. (B)88
  3. (C)80
  4. (D)66

Correct answer: (A)

Step-by-step solution →
Q87·MathematicsSingle correctJEE Main 2024
If ∫0113+x+1+x dx=a+b2+c3\displaystyle\int_0^1\dfrac{1}{\sqrt{3+x}+\sqrt{1+x}}\,dx=a+b\sqrt2+c\sqrt3∫01​3+x​+1+x​1​dx=a+b2​+c3​, where aaa, bbb, ccc are rational numbers, then 2a+3b−4c2a+3b-4c2a+3b−4c is equal to:
  1. (A)4
  2. (B)10
  3. (C)7
  4. (D)8

Correct answer: (D)

Step-by-step solution →
Q88·MathematicsNumericalJEE Main 2024
Let f(x)=∫0xg(t)log⁡e(1−t1+t)dtf(x)=\displaystyle\int_{0}^{x} g(t)\log_e\left(\dfrac{1-t}{1+t}\right)dtf(x)=∫0x​g(t)loge​(1+t1−t​)dt, where ggg is a continuous odd function. If ∫−π/2π/2(f(x)+x2cos⁡x1+ex)dx=(πα)2−α\displaystyle\int_{-\pi/2}^{\pi/2}\left(f(x)+\dfrac{x^2\cos x}{1+e^x}\right)dx=\left(\dfrac{\pi}{\alpha}\right)^2-\alpha∫−π/2π/2​(f(x)+1+exx2cosx​)dx=(απ​)2−α, then α\alphaα is equal to __________.

Correct answer: 2

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Q89·MathematicsIntegerJEE Advanced 2023
For x∈Rx \in \mathbb{R}x∈R, let tan⁡−1(x)∈(−π2,π2)\tan^{-1}(x) \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)tan−1(x)∈(−2π​,2π​). Then the minimum value of the function f:R→Rf : \mathbb{R} \to \mathbb{R}f:R→R defined by f(x)=∫0xtan⁡−1xe(t−cos⁡t)1+t2023 dtf(x) = \int_{0}^{x \tan^{-1} x} \frac{e^{(t - \cos t)}}{1 + t^{2023}}\,dtf(x)=∫0xtan−1x​1+t2023e(t−cost)​dt is

Correct answer: 0

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Q90·MathematicsSingle correctJEE Main 2023
If ∫011(5+2x−2x2)(1+e(2−4x)) dx=1αlog⁡e(α+1β)\int_0^1 \frac{1}{\left(5 + 2x - 2x^2\right)\left(1 + e^{(2-4x)}\right)}\,dx = \frac{1}{\alpha}\log_e\left(\frac{\alpha + 1}{\beta}\right)∫01​(5+2x−2x2)(1+e(2−4x))1​dx=α1​loge​(βα+1​), α,β>0\alpha, \beta > 0α,β>0, then α4−β4\alpha^4 - \beta^4α4−β4 is equal to:
  1. (A)212121
  2. (B)000
  3. (C)191919
  4. (D)−21-21−21

Correct answer: (A)

Step-by-step solution →
Q91·MathematicsNumericalJEE Main 2023
Let for x∈R, S0(x)=x, Sk(x)=Ck+k∫0xSk−1(t)dtx\in\mathbb{R},\,S_{0}(x)=x,\,S_{k}(x)=C_{k}+k\displaystyle\int_{0}^{x}S_{k-1}(t)dtx∈R,S0​(x)=x,Sk​(x)=Ck​+k∫0x​Sk−1​(t)dt, where Ck=1−∫01Sk−1(x)dx, k=1,2,3,…C_{k}=1-\displaystyle\int_{0}^{1}S_{k-1}(x)dx,\,k=1,2,3,\dotsCk​=1−∫01​Sk−1​(x)dx,k=1,2,3,…. Then S2(3)+6C3S_{2}(3)+6C_{3}S2​(3)+6C3​ is equal to _____.

Correct answer: 18

Step-by-step solution →
Q92·MathematicsNumericalJEE Main 2023
Let fn=∫0π2(∑k=1nsin⁡k−1x)(∑k=1n(2k−1)sin⁡k−1x)cos⁡x dxf_n = \int_{0}^{\frac{\pi}{2}} \left(\sum\limits_{k=1}^{n} \sin^{k-1}x\right)\left(\sum\limits_{k=1}^{n} (2k-1)\sin^{k-1}x\right)\cos x\, dxfn​=∫02π​​(k=1∑n​sink−1x)(k=1∑n​(2k−1)sink−1x)cosxdx, n∈Nn \in \mathbb{N}n∈N. Then f21−f20f_{21} - f_{20}f21​−f20​ is equal to

Correct answer: 41

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Q93·MathematicsSingle correctJEE Main 2023
The value of ∫0π4e−x+∫0π4e−xtan⁡50x dx∫0π4e−x(tan⁡49x+tan⁡51x)dx\dfrac{\displaystyle\int_{0}^{\frac{\pi}{4}} e^{-x} + \int_{0}^{\frac{\pi}{4}} e^{-x}\tan^{50}x\,dx}{\displaystyle\int_{0}^{\frac{\pi}{4}} e^{-x}\left(\tan^{49}x + \tan^{51}x\right)dx}∫04π​​e−x(tan49x+tan51x)dx∫04π​​e−x+∫04π​​e−xtan50xdx​ is
  1. (A)50
  2. (B)49
  3. (C)51
  4. (D)52

Correct answer: (A)

Step-by-step solution →
Q94·MathematicsSingle correctJEE Main 2023
∫026e3x+6e2x+11ex+6 dx\displaystyle\int_{0}^{2}\dfrac{6}{e^{3x}+6e^{2x}+11e^{x}+6}\,dx∫02​e3x+6e2x+11ex+66​dx is equal to:
  1. (A)log⁡e51281\log_e\dfrac{512}{81}loge​81512​
  2. (B)log⁡e3227\log_e\dfrac{32}{27}loge​2732​
  3. (C)log⁡e25681\log_e\dfrac{256}{81}loge​81256​
  4. (D)log⁡e6427\log_e\dfrac{64}{27}loge​2764​

Correct answer: (B)

Step-by-step solution →
Q95·MathematicsNumericalJEE Main 2023
If ∫−0.150.15∣100x2−1∣ dx=k3000\int_{-0.15}^{0.15}\left|100x^2-1\right|\,dx=\dfrac{k}{3000}∫−0.150.15​​100x2−1​dx=3000k​, then k is equal to _________.

Correct answer: 575

Step-by-step solution →
Q96·MathematicsSingle correctJEE Main 2023
The value of the integral ∫−log⁡e2log⁡e2ex(log⁡e(ex+1+e2x))dx\int_{-\log_e 2}^{\log_e 2} e^x \left( \log_e \left( e^x + \sqrt{1 + e^{2x}} \right) \right) dx∫−loge​2loge​2​ex(loge​(ex+1+e2x​))dx is equal to
  1. (A)log⁡e(2(2+5)1+5)−52\log_e \left( \frac{2(2 + \sqrt{5})}{\sqrt{1 + \sqrt{5}}} \right) - \frac{\sqrt{5}}{2}loge​(1+5​​2(2+5​)​)−25​​
  2. (B)log⁡e(2(3−5)21+5)+52\log_e \left( \frac{\sqrt{2}(3 - \sqrt{5})^2}{\sqrt{1 + \sqrt{5}}} \right) + \frac{\sqrt{5}}{2}loge​(1+5​​2​(3−5​)2​)+25​​
  3. (C)log⁡e((2+5)21+5)+52\log_e \left( \frac{(2 + \sqrt{5})^2}{\sqrt{1 + \sqrt{5}}} \right) + \frac{\sqrt{5}}{2}loge​(1+5​​(2+5​)2​)+25​​
  4. (D)log⁡e(2(2+5)21+5)−52\log_e \left( \frac{\sqrt{2}(2 + \sqrt{5})^2}{\sqrt{1 + \sqrt{5}}} \right) - \frac{\sqrt{5}}{2}loge​(1+5​​2​(2+5​)2​)−25​​

Correct answer: (D)

Step-by-step solution →
Q97·MathematicsSingle correctJEE Main 2023
Let the function f:[0,2]→Rf:[0,2]\to\mathbb{R}f:[0,2]→R be defined as f(x)={emin⁡{x2, x−[x]},x∈[0,1)e[x−log⁡ex],x∈[1,2]f(x)=\begin{cases} e^{\min\{x^2,\,x-[x]\}}, & x\in[0,1) \\ e^{[x-\log_e x]}, & x\in[1,2] \end{cases}f(x)={emin{x2,x−[x]},e[x−loge​x],​x∈[0,1)x∈[1,2]​ where [t][t][t] denotes the greatest integer less than or equal to ttt. Then the value of the integral ∫02x f(x) dx\int_0^2 x\,f(x)\,dx∫02​xf(x)dx is
  1. (A)2e−12e-12e−1
  2. (B)2e+3e22e+\dfrac{3e}{2}2e+23e​
  3. (C)2e−122e-\dfrac{1}{2}2e−21​
  4. (D)(e−1)(e2+12)(e-1)\left(e^2+\dfrac{1}{2}\right)(e−1)(e2+21​)

Correct answer: (C)

Step-by-step solution →
Q98·MathematicsNumericalJEE Main 2023
For m,n>0m, n > 0m,n>0, let α(m,n)=∫02tm(1+3t)n dt\alpha(m, n) = \int_0^2 t^m (1 + 3t)^n \, dtα(m,n)=∫02​tm(1+3t)ndt. If 11α(10,6)+18α(11,5)=p(14)611\alpha(10, 6) + 18\alpha(11, 5) = p(14)^611α(10,6)+18α(11,5)=p(14)6, then ppp is equal to _______ .

Correct answer: 32

Step-by-step solution →
Q99·MathematicsSingle correctJEE Main 2023
If f:R→Rf:\mathbb{R}\to\mathbb{R}f:R→R be a continuous function satisfying ∫0π/2f(sin⁡2x)sin⁡x dx+α∫0π/4f(cos⁡2x)cos⁡x dx=0\displaystyle\int_0^{\pi/2} f(\sin 2x)\sin x\,dx+\alpha\int_0^{\pi/4} f(\cos 2x)\cos x\,dx=0∫0π/2​f(sin2x)sinxdx+α∫0π/4​f(cos2x)cosxdx=0, then α\alphaα is equal to
  1. (A)−3-\sqrt{3}−3​
  2. (B)2\sqrt{2}2​
  3. (C)3\sqrt{3}3​
  4. (D)−2-\sqrt{2}−2​

Correct answer: (D)

Step-by-step solution →
Q100·MathematicsSingle correctJEE Main 2023
Let fff be a continuous function satisfying ∫0t2(f(x)+x2)dx=43t3, ∀ t>0\int_0^{t^2}\left(f(x)+x^2\right)dx=\frac{4}{3}t^3,\ \forall\ t>0∫0t2​(f(x)+x2)dx=34​t3, ∀ t>0. Then f(π24)f\left(\frac{\pi^2}{4}\right)f(4π2​) is equal to:
  1. (A)π(1−π316)\pi\left(1-\frac{\pi^3}{16}\right)π(1−16π3​)
  2. (B)−π2(1+π216)-\pi^2\left(1+\frac{\pi^2}{16}\right)−π2(1+16π2​)
  3. (C)−π(1+π216)-\pi\left(1+\frac{\pi^2}{16}\right)−π(1+16π2​)
  4. (D)π2(1−π216)\pi^2\left(1-\frac{\pi^2}{16}\right)π2(1−16π2​)

Correct answer: (A)

Step-by-step solution →
Q101·MathematicsSingle correctJEE Main 2023
Let fff be a differentiable function such that x2f(x)−x=4∫0xt f(t) dtx^2 f(x)-x=4\int_0^x t\,f(t)\,dtx2f(x)−x=4∫0x​tf(t)dt, f(1)=23f(1)=\dfrac23f(1)=32​. Then 18 f(3)18\,f(3)18f(3) is equal to
  1. (A)160160160
  2. (B)210210210
  3. (C)180180180
  4. (D)150150150

Correct answer: (A)

Step-by-step solution →
Q102·MathematicsNumericalJEE Main 2023
Let [t][t][t] denotes the greatest integer ≤t\le t≤t. Then 2π∫π/65π/6(8[cosec⁡x]−5[cot⁡x])dx\frac{2}{\pi}\int_{\pi/6}^{5\pi/6}\left(8[\operatorname{cosec} x]-5[\cot x]\right)dxπ2​∫π/65π/6​(8[cosecx]−5[cotx])dx is equal to

Correct answer: 14

Step-by-step solution →
Q103·MathematicsNumericalJEE Main 2023
Let [t][t][t] denote the greatest integer function. If ∫02.4[x2]dx=α+β2+γ3+δ5\displaystyle\int_0^{2.4}\left[x^{2}\right]dx=\alpha+\beta\sqrt{2}+\gamma\sqrt{3}+\delta\sqrt{5}∫02.4​[x2]dx=α+β2​+γ3​+δ5​, then α+β+γ+δ\alpha+\beta+\gamma+\deltaα+β+γ+δ is equal to

Correct answer: 6

Step-by-step solution →
Q104·MathematicsSingle correctJEE Main 2023
Let 5f(x)+4f(1x)=1x+3, x>05f(x)+4f\left(\dfrac{1}{x}\right)=\dfrac{1}{x}+3,\,x>05f(x)+4f(x1​)=x1​+3,x>0. Then 18∫12f(x) dx18\displaystyle\int_{1}^{2}f(x)\,dx18∫12​f(x)dx is equal to:
  1. (A)10log⁡e2−610\log_e 2-610loge​2−6
  2. (B)10log⁡e2+610\log_e 2+610loge​2+6
  3. (C)6log⁡e2+36\log_e 2+36loge​2+3
  4. (D)6log⁡e2−36\log_e 2-36loge​2−3

Correct answer: (A)

Step-by-step solution →
Q105·MathematicsNumericalJEE Main 2023
Let f(x)=x(1+xn)1nf(x)=\frac{x}{\left(1+x^n\right)^{\frac{1}{n}}}f(x)=(1+xn)n1​x​, x∈R−{−1}x\in \mathbb{R}-\{-1\}x∈R−{−1}, n∈Nn\in \mathbb{N}n∈N, n>2n>2n>2. If fn(x)=(f∘f∘⋯∘f⏟n times)(x)f^n(x)=(\underbrace{f\circ f\circ \cdots \circ f}_{n\text{ times}})(x)fn(x)=(n timesf∘f∘⋯∘f​​)(x), then lim⁡n→∞∫xn−2(fn(x))dx\lim_{n\to\infty}\int x^{n-2}\left(f^n(x)\right)dxlimn→∞​∫xn−2(fn(x))dx is equal to

Correct answer: 0

Step-by-step solution →
Q106·MathematicsSingle correctJEE Main 2023
Let f(x)f(x)f(x) be a function satisfying f(x)+f(π−x)=π2f(x)+f(\pi-x)=\pi^2f(x)+f(π−x)=π2, ∀x∈R\forall x\in \mathbb{R}∀x∈R. Then ∫0πf(x)sin⁡x dx\int_0^\pi f(x)\sin x\,dx∫0π​f(x)sinxdx is equal to
  1. (A)π24\frac{\pi^2}{4}4π2​
  2. (B)π22\frac{\pi^2}{2}2π2​
  3. (C)2π22\pi^22π2
  4. (D)π2\pi^2π2

Correct answer: (D)

Step-by-step solution →
Q107·MathematicsSingle correctJEE Main 2023
lim⁡n→∞[11+n+12+n+13+n+⋯+12n]\lim_{n\to\infty}\left[\frac{1}{1+n}+\frac{1}{2+n}+\frac{1}{3+n}+\cdots+\frac{1}{2n}\right]limn→∞​[1+n1​+2+n1​+3+n1​+⋯+2n1​] is equal to
  1. (A)log⁡e2\log_e 2loge​2
  2. (B)log⁡e ⁣(23)\log_e\!\left(\frac{2}{3}\right)loge​(32​)
  3. (C)log⁡e ⁣(32)\log_e\!\left(\frac{3}{2}\right)loge​(23​)
  4. (D)000

Correct answer: (A)

Step-by-step solution →
Q108·MathematicsNumericalJEE Main 2023
If ∫01(x21+x14+x7)(2x14+3x7+6)1/7 dx=1l(11)m/n\int_0^1 (x^{21}+x^{14}+x^7)(2x^{14}+3x^7+6)^{1/7}\,dx=\frac{1}{l}(11)^{m/n}∫01​(x21+x14+x7)(2x14+3x7+6)1/7dx=l1​(11)m/n where l,m,n∈Nl,m,n\in\mathbb{N}l,m,n∈N, mmm and nnn are coprime, then l+m+nl+m+nl+m+n is equal to _______ .

Correct answer: 63

Step-by-step solution →
Q109·MathematicsNumericalJEE Main 2023
If ∫−π2π25cos⁡x(1+cos⁡xcos⁡3x+cos⁡2x+cos⁡3xcos⁡3x)1+5cos⁡xdx=kπ16\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\dfrac{5^{\cos x}\left(1+\cos x\cos 3x+\cos^2 x+\cos^3 x\cos 3x\right)}{1+5^{\cos x}}dx=\dfrac{k\pi}{16}∫−2π​2π​​1+5cosx5cosx(1+cosxcos3x+cos2x+cos3xcos3x)​dx=16kπ​, then k is equal to

Correct answer: 13

Step-by-step solution →
Q110·MathematicsSingle correctJEE Main 2023
The value of the integral ∫−π4π4x+π42−cos⁡2xdx\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\dfrac{x+\frac{\pi}{4}}{2-\cos 2x}dx∫−4π​4π​​2−cos2xx+4π​​dx is:
  1. (A)π2123\dfrac{\pi^2}{12\sqrt3}123​π2​
  2. (B)π263\dfrac{\pi^2}{6\sqrt3}63​π2​
  3. (C)π26\dfrac{\pi^2}{6}6π2​
  4. (D)π233\dfrac{\pi^2}{3\sqrt3}33​π2​

Correct answer: (B)

Step-by-step solution →
Q111·MathematicsSingle correctJEE Main 2023
If ϕ(x)=1x∫π4x(42sin⁡t−3ϕ′(t))dt, x>0\phi(x)=\dfrac{1}{\sqrt{x}}\displaystyle\int_{\frac{\pi}{4}}^{x}\left(4\sqrt2\sin t-3\phi'(t)\right)dt,\ x>0ϕ(x)=x​1​∫4π​x​(42​sint−3ϕ′(t))dt, x>0, then ϕ′(π4)\phi'\left(\dfrac{\pi}{4}\right)ϕ′(4π​) is equal to:
  1. (A)86+π\dfrac{8}{6+\sqrt\pi}6+π​8​
  2. (B)46+π\dfrac{4}{6+\sqrt\pi}6+π​4​
  3. (C)8π\dfrac{8}{\sqrt\pi}π​8​
  4. (D)46−π\dfrac{4}{6-\sqrt\pi}6−π​4​

Correct answer: (A)

Step-by-step solution →
Q112·MathematicsSingle correctJEE Main 2023
The value of ∫π/3π/2(2+3sin⁡x)sin⁡x(1+cos⁡x) dx\int_{\pi/3}^{\pi/2}\frac{(2+3\sin x)}{\sin x(1+\cos x)}\,dx∫π/3π/2​sinx(1+cosx)(2+3sinx)​dx is equal to
  1. (A)103−3−log⁡e3\frac{10}{3}-\sqrt{3}-\log_{e}\sqrt{3}310​−3​−loge​3​
  2. (B)72−3−log⁡e3\frac{7}{2}-\sqrt{3}-\log_{e}\sqrt{3}27​−3​−loge​3​
  3. (C)−2+33+log⁡e3-2+3\sqrt{3}+\log_{e}\sqrt{3}−2+33​+loge​3​
  4. (D)103−3+log⁡e3\frac{10}{3}-\sqrt{3}+\log_{e}\sqrt{3}310​−3​+loge​3​

Correct answer: (D)

Step-by-step solution →
Q113·MathematicsSingle correctJEE Main 2023
Let α>0\alpha>0α>0. If ∫0αxx+α−xdx=16+20215\displaystyle\int_{0}^{\alpha}\dfrac{x}{\sqrt{x+\alpha}-\sqrt{x}}dx=\dfrac{16+20\sqrt2}{15}∫0α​x+α​−x​x​dx=1516+202​​, then α\alphaα is equal to:
  1. (A)444
  2. (B)222\sqrt222​
  3. (C)2\sqrt22​
  4. (D)222

Correct answer: (D)

Step-by-step solution →
Q114·MathematicsSingle correctJEE Main 2023
If [t][t][t] denotes the greatest integer ≤t\leq t≤t, then the value of 3(e−1)e∫12x2e[x3]+[x4] dx\dfrac{3(e-1)}{e}\int_1^2 x^2 e^{[x^3]+[x^4]}\,dxe3(e−1)​∫12​x2e[x3]+[x4]dx is:
  1. (A)e8−1e^8-1e8−1
  2. (B)e7−1e^7-1e7−1
  3. (C)e8−ee^8-ee8−e
  4. (D)e9−ee^9-ee9−e

Correct answer: (C)

Step-by-step solution →
Q115·MathematicsSingle correctJEE Main 2023
lim⁡n→∞3n{4+(2+1n)2+(2+2n)2+⋯+(3−1n)2}\displaystyle\lim_{n \to \infty} \dfrac{3}{n}\left\{4 + \left(2 + \dfrac{1}{n}\right)^2 + \left(2 + \dfrac{2}{n}\right)^2 + \cdots + \left(3 - \dfrac{1}{n}\right)^2\right\}n→∞lim​n3​{4+(2+n1​)2+(2+n2​)2+⋯+(3−n1​)2} is equal to:
  1. (A)121212
  2. (B)193\dfrac{19}{3}319​
  3. (C)000
  4. (D)191919

Correct answer: (D)

Step-by-step solution →
Q116·MathematicsSingle correctJEE Main 2023
The value of the integral ∫1/22tan⁡−1xxdx\displaystyle\int_{1/2}^{2}\dfrac{\tan^{-1}x}{x}dx∫1/22​xtan−1x​dx is equal to:
  1. (A)π2log⁡e2\dfrac{\pi}{2}\log_e 22π​loge​2
  2. (B)πlog⁡e2\pi\log_e 2πloge​2
  3. (C)12log⁡e2\dfrac12\log_e 221​loge​2
  4. (D)π4log⁡e2\dfrac{\pi}{4}\log_e 24π​loge​2

Correct answer: (A)

Step-by-step solution →
Q117·MathematicsSingle correctJEE Main 2023
Let [x][x][x] denote the greatest integer ≤x\leq x≤x. Consider the function f(x)=max⁡{x2,1+[x]}f(x) = \max\{x^2, 1 + [x]\}f(x)=max{x2,1+[x]}. Then the value of the integral ∫02f(x) dx\int_0^2 f(x)\,dx∫02​f(x)dx is
  1. (A)5+423\dfrac{5 + 4\sqrt{2}}{3}35+42​​
  2. (B)4+523\dfrac{4 + 5\sqrt{2}}{3}34+52​​
  3. (C)1+523\dfrac{1 + 5\sqrt{2}}{3}31+52​​
  4. (D)8+423\dfrac{8 + 4\sqrt{2}}{3}38+42​​

Correct answer: (A)

Step-by-step solution →
Q118·MathematicsSingle correctJEE Main 2023
Let f(x)=x+aπ2−4sin⁡x+bπ2−4cos⁡xf(x) = x + \dfrac{a}{\pi^2 - 4}\sin x + \dfrac{b}{\pi^2 - 4}\cos xf(x)=x+π2−4a​sinx+π2−4b​cosx, x∈Rx \in \mathbb{R}x∈R be a function which satisfies f(x)=x+∫0π/2sin⁡(x+y)f(y) dyf(x) = x + \int_0^{\pi/2} \sin(x + y) f(y)\,dyf(x)=x+∫0π/2​sin(x+y)f(y)dy. Then (a+b)(a + b)(a+b) is equal to
  1. (A)−2π(π−2)-2\pi(\pi - 2)−2π(π−2)
  2. (B)−2π(π+2)-2\pi(\pi + 2)−2π(π+2)
  3. (C)−π(π−2)-\pi(\pi - 2)−π(π−2)
  4. (D)−π(π+2)-\pi(\pi + 2)−π(π+2)

Correct answer: (B)

Step-by-step solution →
Q119·MathematicsSingle correctJEE Main 2023
The value of the integral ∫12(t4+1t6+1)dt\displaystyle\int_{1}^{2}\left(\dfrac{t^{4}+1}{t^{6}+1}\right)dt∫12​(t6+1t4+1​)dt is:
  1. (A)tan⁡−12−13tan⁡−18+π4\tan^{-1}2-\dfrac13\tan^{-1}8+\dfrac{\pi}{4}tan−12−31​tan−18+4π​
  2. (B)tan⁡−112+13tan⁡−18−π3\tan^{-1}\dfrac12+\dfrac13\tan^{-1}8-\dfrac{\pi}{3}tan−121​+31​tan−18−3π​
  3. (C)tan⁡−112−13tan⁡−18+π3\tan^{-1}\dfrac12-\dfrac13\tan^{-1}8+\dfrac{\pi}{3}tan−121​−31​tan−18+3π​
  4. (D)tan⁡−12+13tan⁡−18−π3\tan^{-1}2+\dfrac13\tan^{-1}8-\dfrac{\pi}{3}tan−12+31​tan−18−3π​

Correct answer: (D)

Step-by-step solution →
Q120·MathematicsSingle correctJEE Main 2023
The minimum value of the function f(x)=∫02e∣x−t∣ dtf(x)=\displaystyle\int_0^2 e^{|x-t|}\,dtf(x)=∫02​e∣x−t∣dt is:
  1. (A)e(e−1)e(e-1)e(e−1)
  2. (B)2(e−1)2(e-1)2(e−1)
  3. (C)222
  4. (D)2e−12e-12e−1

Correct answer: (B)

Step-by-step solution →
Q121·MathematicsNumericalJEE Main 2023
If ∫1/33∣log⁡ex∣ dx=mnlog⁡e(n2e)\displaystyle\int_{1/3}^{3}|\log_{e}x|\,dx=\dfrac{m}{n}\log_{e}\left(\dfrac{n^{2}}{e}\right)∫1/33​∣loge​x∣dx=nm​loge​(en2​), where mmm and nnn are coprime natural numbers, then m2+n2−5m^{2}+n^{2}-5m2+n2−5 is equal to

Correct answer: 20

Step-by-step solution →
Q122·MathematicsSingle correctJEE Main 2023
The integral 16∫12dxx3(x2+2)216\displaystyle\int_{1}^{2} \dfrac{dx}{x^3 (x^2 + 2)^2}16∫12​x3(x2+2)2dx​ is equal to
  1. (A)1112−log⁡e4\dfrac{11}{12} - \log_e 41211​−loge​4
  2. (B)116−log⁡e4\dfrac{11}{6} - \log_e 4611​−loge​4
  3. (C)116+log⁡e4\dfrac{11}{6} + \log_e 4611​+loge​4
  4. (D)1112+log⁡e4\dfrac{11}{12} + \log_e 41211​+loge​4

Correct answer: (B)

Step-by-step solution →
Q123·MathematicsNumericalJEE Main 2023
The value of 12∫03∣x2−3x+2∣ dx12\displaystyle\int_0^3 |x^2-3x+2|\,dx12∫03​∣x2−3x+2∣dx is _______ .

Correct answer: 22

Step-by-step solution →
Q124·MathematicsNumericalJEE Main 2023
The value of 8π∫0π/2(cos⁡x)2023(sin⁡x)2023+(cos⁡x)2023 dx\dfrac{8}{\pi}\displaystyle\int_0^{\pi/2}\dfrac{(\cos x)^{2023}}{(\sin x)^{2023}+(\cos x)^{2023}}\,dxπ8​∫0π/2​(sinx)2023+(cosx)2023(cosx)2023​dx is _______ .

Correct answer: 2

Step-by-step solution →
Q125·MathematicsSingle correctJEE Main 2023
∫324334489−4x2 dx\displaystyle\int_{\frac{3\sqrt2}{4}}^{\frac{3\sqrt3}{4}}\dfrac{48}{\sqrt{9-4x^2}}\,dx∫432​​433​​​9−4x2​48​dx is equal to
  1. (A)2π2\pi2π
  2. (B)π6\dfrac{\pi}{6}6π​
  3. (C)π3\dfrac{\pi}{3}3π​
  4. (D)π2\dfrac{\pi}{2}2π​

Correct answer: (A)

Step-by-step solution →
Q126·MathematicsIntegerJEE Advanced 2022
The greatest integer less than or equal to ∫12log⁡2(x3+1)dx+∫1log⁡29(2x−1)1/3dx\int_{1}^{2} \log_{2}\left(x^{3} + 1\right)dx + \int_{1}^{\log_{2}9} \left(2^{x} - 1\right)^{1/3} dx∫12​log2​(x3+1)dx+∫1log2​9​(2x−1)1/3dx is ______.

Correct answer: 5

Step-by-step solution →
Q127·MathematicsSingle correctJEE Advanced 2022
For positive integer n, define f(n) = n + 16+5n−3n24n+3n2\frac{16+5n-3n^{2}}{4n+3n^{2}}4n+3n216+5n−3n2​ + 32+n−3n28n+3n2\frac{32+n-3n^{2}}{8n+3n^{2}}8n+3n232+n−3n2​ + 48−3n−3n212n+3n2\frac{48-3n-3n^{2}}{12n+3n^{2}}12n+3n248−3n−3n2​ + … + 25n−7n27n2\frac{25n-7n^{2}}{7n^{2}}7n225n−7n2​ . Then, the value of lim⁡n→∞\lim_{n \to \infty}limn→∞​ f(n) is equal to
  1. (A)3 + 43\frac{4}{3}34​loge_{e}e​7
  2. (B)4 − 34\frac{3}{4}43​loge_{e}e​(73)\left(\frac{7}{3}\right)(37​)
  3. (C)4 − 43\frac{4}{3}34​loge_{e}e​(73)\left(\frac{7}{3}\right)(37​)
  4. (D)3 + 34\frac{3}{4}43​loge_{e}e​7 .

Correct answer: (B)

Step-by-step solution →
Q128·MathematicsMultiple correctJEE Advanced 2022
Consider the equation ∫1e(log⁡ex)12x(a−(log⁡ex)32)2 dx=1\int_{1}^{e} \frac{\left(\log_{e} x\right)^{\frac{1}{2}}}{x\left(a - \left(\log_{e} x\right)^{\frac{3}{2}}\right)^{2}} \, dx = 1∫1e​x(a−(loge​x)23​)2(loge​x)21​​dx=1, a∈(−∞,0)∪(1,∞)a \in \left(-\infty, 0\right) \cup \left(1, \infty\right)a∈(−∞,0)∪(1,∞). Which of the following statements is/are TRUE?
  1. (A)No aaa satisfies the above equation
  2. (B)An integer aaa satisfies the above equation
  3. (C)An irrational number aaa satisfies the above equation
  4. (D)More than one aaa satisfy the above equation

Correct answer: (C), (D)

Step-by-step solution →
Q129·MathematicsSingle correctJEE Main 2022
The integral ∫0π213+2sin⁡x+cos⁡x dx\int_{0}^{\frac{\pi}{2}} \frac{1}{3 + 2\sin x + \cos x}\,dx∫02π​​3+2sinx+cosx1​dx is equal to:
  1. (A)tan⁡−1(2)\tan^{-1}(2)tan−1(2)
  2. (B)tan⁡−1(2)−π4\tan^{-1}(2) - \frac{\pi}{4}tan−1(2)−4π​
  3. (C)12tan⁡−1(2)−π8\frac{1}{2}\tan^{-1}(2) - \frac{\pi}{8}21​tan−1(2)−8π​
  4. (D)12\frac{1}{2}21​

Correct answer: (B)

Step-by-step solution →
Q130·MathematicsSingle correctJEE Main 2022
If [t][t][t] denotes the greatest integer ≤t\leq t≤t, then the value of ∫01[2x−∣3x2−5x+2∣+1]dx\int_{0}^{1} \left[ 2x - |3x^2 - 5x + 2| + 1 \right] dx∫01​[2x−∣3x2−5x+2∣+1]dx is:
  1. (A)37+13−46\frac{\sqrt{37} + \sqrt{13} - 4}{6}637​+13​−4​
  2. (B)37−13−46\frac{\sqrt{37} - \sqrt{13} - 4}{6}637​−13​−4​
  3. (C)−37−13+46\frac{-\sqrt{37} - \sqrt{13} + 4}{6}6−37​−13​+4​
  4. (D)−37+13+46\frac{-\sqrt{37} + \sqrt{13} + 4}{6}6−37​+13​+4​

Correct answer: (A)

Step-by-step solution →
Q131·MathematicsSingle correctJEE Main 2022
If f(α)=∫1αlog⁡10t1+t dt,α>0f(\alpha) = \int_{1}^{\alpha} \frac{\log_{10} t}{1 + t}\,dt, \alpha > 0f(α)=∫1α​1+tlog10​t​dt,α>0, then f(e3)+f(e−3)f(e^{3}) + f(e^{-3})f(e3)+f(e−3) is equal to :
  1. (A)9
  2. (B)92\frac{9}{2}29​
  3. (C)9log⁡e(10)\frac{9}{\log_{e}(10)}loge​(10)9​
  4. (D)92log⁡e(10)\frac{9}{2\log_{e}(10)}2loge​(10)9​

Correct answer: (D)

Step-by-step solution →
Q132·MathematicsSingle correctJEE Main 2022
Let In(x)=∫0x1(t2+5)ndtI_{n}(x)=\int_{0}^{x}\frac{1}{\left(t^{2}+5\right)^{n}}dtIn​(x)=∫0x​(t2+5)n1​dt, n=1,2,3,….n=1,2,3,\ldots.n=1,2,3,…. Then
  1. (A)50I6−9I5=xI5′50I_{6}-9I_{5}=xI_{5}'50I6​−9I5​=xI5′​
  2. (B)50I6−11I5=xI5′50I_{6}-11I_{5}=xI_{5}'50I6​−11I5​=xI5′​
  3. (C)50I6−9I5=I5′50I_{6}-9I_{5}=I_{5}'50I6​−9I5​=I5′​
  4. (D)50I6−11I5=I5′50I_{6}-11I_{5}=I_{5}'50I6​−11I5​=I5′​

Correct answer: (A)

Step-by-step solution →
Q133·MathematicsSingle correctJEE Main 2022
The minimum value of the twice differentiable function f(x)=∫0xex−tf′(t) dt−(x2−x+1)exf(x) = \int_{0}^{x} e^{x-t} f'(t)\,dt - (x^{2} - x + 1)e^{x}f(x)=∫0x​ex−tf′(t)dt−(x2−x+1)ex, x∈Rx \in Rx∈R, is :
  1. (A)−2e-\frac{2}{\sqrt{e}}−e​2​
  2. (B)−2e-2\sqrt{e}−2e​
  3. (C)−e-\sqrt{e}−e​
  4. (D)2e\frac{2}{\sqrt{e}}e​2​

Correct answer: (A)

Step-by-step solution →
Q134·MathematicsNumericalJEE Main 2022
The value of the integral ∫0π/260sin⁡(6x)sin⁡xdx\int_{0}^{\pi/2}60\frac{\sin(6x)}{\sin x}dx∫0π/2​60sinxsin(6x)​dx is equal to

Correct answer: 104

Step-by-step solution →
Q135·MathematicsNumericalJEE Main 2022
If ∫0315x31+x2+(1+x2)3 dx=α2+β3\int_{0}^{\sqrt{3}} \frac{15x^{3}}{\sqrt{1 + x^{2} + \sqrt{\left(1 + x^{2}\right)^{3}}}}\,dx = \alpha\sqrt{2} + \beta\sqrt{3}∫03​​1+x2+(1+x2)3​​15x3​dx=α2​+β3​, where α\alphaα, β\betaβ are integers, then α+β\alpha + \betaα+β is equal to ______.

Correct answer: 10

Step-by-step solution →
Q136·MathematicsSingle correctJEE Main 2022
I=∫π/4π/3(8sin⁡x−sin⁡2xx)dxI = \int_{\pi/4}^{\pi/3}\left(\frac{8\sin x - \sin 2x}{x}\right)dxI=∫π/4π/3​(x8sinx−sin2x​)dx. Then
  1. (A)π2<I<3π4\frac{\pi}{2} < I < \frac{3\pi}{4}2π​<I<43π​
  2. (B)π5<I<5π12\frac{\pi}{5} < I < \frac{5\pi}{12}5π​<I<125π​
  3. (C)5π12<I<23π\frac{5\pi}{12} < I < \frac{\sqrt{2}}{3}\pi125π​<I<32​​π
  4. (D)3π4<I<π\frac{3\pi}{4} < I < \pi43π​<I<π

Correct answer: (C)

Step-by-step solution →
Q137·MathematicsNumericalJEE Main 2022
Let f(x)=min⁡{[x−1],[x−2],...,[x−10]}f(x) = \min\{[x-1], [x-2], ..., [x-10]\}f(x)=min{[x−1],[x−2],...,[x−10]} where [t][t][t] denotes the greatest integer ≤t\le t≤t. Then ∫010f(x)dx+∫010(f(x))2dx+∫010∣f(x)∣dx\int_{0}^{10} f(x)dx + \int_{0}^{10} (f(x))^{2}dx + \int_{0}^{10} |f(x)|dx∫010​f(x)dx+∫010​(f(x))2dx+∫010​∣f(x)∣dx is equal to ___.

Correct answer: 385

Step-by-step solution →
Q138·MathematicsNumericalJEE Main 2022
Let f be a differentiable function satisfying f(x)=23∫03f(λ2x3)dλf(x) = \frac{2}{\sqrt{3}}\int_{0}^{\sqrt{3}} f\left(\frac{\lambda^{2}x}{3}\right)d\lambdaf(x)=3​2​∫03​​f(3λ2x​)dλ, x>0x > 0x>0 and f(1)=3f(1) = \sqrt{3}f(1)=3​. If y=f(x)y = f(x)y=f(x) passes through the point (α,6)(\alpha, 6)(α,6), then α\alphaα is equal to _______.

Correct answer: 12

Step-by-step solution →
Q139·MathematicsSingle correctJEE Main 2022
Let f:R→Rf : \mathbb{R} \to \mathbb{R}f:R→R be a function defined as f(x)=asin⁡(π[x]2)+[2−x]f(x) = a\sin\left(\frac{\pi[x]}{2}\right) + [2-x]f(x)=asin(2π[x]​)+[2−x], a∈Ra \in \mathbb{R}a∈R, where [t] is the greatest integer less than or equal to ttt. If lim⁡x→−1f(x)\lim_{x \to -1} f(x)limx→−1​f(x) exists, then the value of ∫04f(x) dx\int_{0}^{4} f(x)\,dx∫04​f(x)dx is equal to :
  1. (A)-1
  2. (B)-2
  3. (C)1
  4. (D)2

Correct answer: (B)

Step-by-step solution →
Q140·MathematicsSingle correctJEE Main 2022
∫02(∣2x2−3x∣+[x−12])dx\int_{0}^{2}\left(\left|2x^{2}-3x\right|+\left[x-\frac{1}{2}\right]\right)dx∫02​(​2x2−3x​+[x−21​])dx, where [t][t][t] is the greatest integer function, is equal to:
  1. (A)76\frac{7}{6}67​
  2. (B)1912\frac{19}{12}1219​
  3. (C)3112\frac{31}{12}1231​
  4. (D)32\frac{3}{2}23​

Correct answer: (B)

Step-by-step solution →
Q141·MathematicsSingle correctJEE Main 2022
∫020π(∣sin⁡x∣+∣cos⁡x∣)2dx\int\limits_{0}^{20\pi}\left(\left|\sin x\right|+\left|\cos x\right|\right)^{2}dx0∫20π​(∣sinx∣+∣cosx∣)2dx is equal to :-
  1. (A)10(π+4)10\left(\pi+4\right)10(π+4)
  2. (B)10(π+2)10\left(\pi+2\right)10(π+2)
  3. (C)20(π−2)20\left(\pi-2\right)20(π−2)
  4. (D)20(π+2)20\left(\pi+2\right)20(π+2)

Correct answer: (D)

Step-by-step solution →
Q142·MathematicsNumericalJEE Main 2022
If n(2n+1)∫01(1−xn)2n dx=1177∫01(1−xn)2n+1dxn\left(2n+1\right)\int_{0}^{1}\left(1-x^{n}\right)^{2n}\,dx = 1177\int_{0}^{1}\left(1-x^{n}\right)^{2n+1}dxn(2n+1)∫01​(1−xn)2ndx=1177∫01​(1−xn)2n+1dx, then n ∈ N is equal to _______

Correct answer: 24

Step-by-step solution →
Q143·MathematicsSingle correctJEE Main 2022
If a = lim⁡n→∞∑k=1n2nn2+k2\lim\limits_{n \to \infty}\sum\limits_{k=1}^{n}\dfrac{2n}{n^{2}+k^{2}}n→∞lim​k=1∑n​n2+k22n​ and f(x) = 1−cos⁡x1+cos⁡x\sqrt{\dfrac{1-\cos x}{1+\cos x}}1+cosx1−cosx​​, x ∈ (0,1), then :
  1. (A)22 f(a2)=f ′(a2)2\sqrt{2}\,\mathrm{f}\left(\dfrac{a}{2}\right) = \mathrm{f}\,'\left(\dfrac{a}{2}\right)22​f(2a​)=f′(2a​)
  2. (B)f(a2)f ′(a2)=2\mathrm{f}\left(\dfrac{a}{2}\right)\mathrm{f}\,'\left(\dfrac{a}{2}\right) = \sqrt{2}f(2a​)f′(2a​)=2​
  3. (C)2 f(a2)=f ′(a2)\sqrt{2}\,\mathrm{f}\left(\dfrac{a}{2}\right) = \mathrm{f}\,'\left(\dfrac{a}{2}\right)2​f(2a​)=f′(2a​)
  4. (D)f(a2)=2 f ′(a2)\mathrm{f}\left(\dfrac{a}{2}\right) = \sqrt{2}\,\mathrm{f}\,'\left(\dfrac{a}{2}\right)f(2a​)=2​f′(2a​)

Correct answer: (C)

Step-by-step solution →
Q144·MathematicsSingle correctJEE Main 2022
For any real number x, let [x] denote the largest integer less than equal to x. Let f be a real valued function defined on the interval [−10,10][-10, 10][−10,10] by f(x)={x−[x],if(x) is odd1+[x]−xif(x) is evenf(x) = \begin{cases} x - [x], & \text{if}(x) \text{ is odd} \\ 1 + [x] - x & \text{if}(x) \text{ is even} \end{cases}f(x)={x−[x],1+[x]−x​if(x) is oddif(x) is even​ Then the value of π210∫−1010f(x)cos⁡πx dx\frac{\pi^{2}}{10} \int_{-10}^{10} f(x) \cos \pi x\, dx10π2​∫−1010​f(x)cosπxdx is :
  1. (A)4
  2. (B)2
  3. (C)1
  4. (D)0

Correct answer: (A)

Step-by-step solution →
Q145·MathematicsNumericalJEE Main 2022
If lim⁡n→∞(n+1)k−1nk+1[(nk+1)+(nk+2)+…+(nk+n)]=33.lim⁡n→∞1nk+1⋅[1k+2k+3k+…+nk]\lim_{n\to\infty} \frac{(n+1)^{k-1}}{n^{k+1}} [(nk + 1) + (nk + 2) + \ldots + (nk + n)] = 33. \lim_{n\to\infty} \frac{1}{n^{k+1}} \cdot [1^{k} + 2^{k} + 3^{k} + \ldots + n^{k}]limn→∞​nk+1(n+1)k−1​[(nk+1)+(nk+2)+…+(nk+n)]=33.limn→∞​nk+11​⋅[1k+2k+3k+…+nk], then the integral value of k is equal to ________.

Correct answer: 5

Step-by-step solution →
Q146·MathematicsSingle correctJEE Main 2022
∫05cos⁡(π(x−[x2]))dx\int_{0}^{5} \cos\left(\pi\left(x - \left[\frac{x}{2}\right]\right)\right) dx∫05​cos(π(x−[2x​]))dx, Where [t] denotes greatest integer less than or equal to t, is equal to :
  1. (A)−3
  2. (B)−2
  3. (C)2
  4. (D)0

Correct answer: (D)

Step-by-step solution →
Q147·MathematicsSingle correctJEE Main 2022
Let fff be a real valued continuous function on [0,1][0,1][0,1] and f(x)=x+∫01(x−t)f(t)dtf(x) = x + \int_{0}^{1}(x - t)f(t)dtf(x)=x+∫01​(x−t)f(t)dt. Then which of the following points (x,y) lies on the curve y=f(x)y = f(x)y=f(x)?
  1. (A)(2,4)(2, 4)(2,4)
  2. (B)(1,2)(1, 2)(1,2)
  3. (C)(4,17)(4, 17)(4,17)
  4. (D)(6,8)(6, 8)(6,8)

Correct answer: (D)

Step-by-step solution →
Q148·MathematicsSingle correctJEE Main 2022
Let [t] denote the greatest integer less than or equal to t. Then, the value of the integral ∫01[−8x2+6x−1] dx\int_{0}^{1}[-8x^{2}+6x-1]\,dx∫01​[−8x2+6x−1]dx is equal to
  1. (A)−1-1−1
  2. (B)−54-\frac{5}{4}−45​
  3. (C)17−138\frac{\sqrt{17}-13}{8}817​−13​
  4. (D)17−168\frac{\sqrt{17}-16}{8}817​−16​

Correct answer: (C)

Step-by-step solution →
Q149·MathematicsSingle correctJEE Main 2022
Let f:R→Rf : \mathbb{R} \to \mathbb{R}f:R→R be a differentiable function such that f(π4)=2,f(π2)=0f\left(\frac{\pi}{4}\right) = \sqrt{2}, f\left(\frac{\pi}{2}\right) = 0f(4π​)=2​,f(2π​)=0 and f′(π2)=1f'\left(\frac{\pi}{2}\right) = 1f′(2π​)=1 and let g(x)=∫xπ/4(f′(t)sec⁡t+tan⁡tsec⁡t f(t))dtg(x) = \int_{x}^{\pi/4} \left(f'(t)\sec t + \tan t \sec t\, f(t)\right) dtg(x)=∫xπ/4​(f′(t)sect+tantsectf(t))dt for x∈[π4,π2)x \in \left[\frac{\pi}{4}, \frac{\pi}{2}\right)x∈[4π​,2π​). Then lim⁡x→(π2)−g(x)\lim_{x \to \left(\frac{\pi}{2}\right)^{-}} g(x)limx→(2π​)−​g(x) is equal to
  1. (A)2
  2. (B)3
  3. (C)4
  4. (D)−3-3−3

Correct answer: (B)

Step-by-step solution →
Q150·MathematicsSingle correctJEE Main 2022
Let f:R→Rf : \mathbb{R} \to \mathbb{R}f:R→R be continuous function satisfying f(x)+f(x+k)=nf(x) + f(x + k) = nf(x)+f(x+k)=n, for all x∈Rx \in \mathbb{R}x∈R where k>0k > 0k>0 and n is a positive integer. If I1=∫04nkf(x)dxI_1 = \int_{0}^{4nk} f(x)dxI1​=∫04nk​f(x)dx and I2=∫−k3kf(x)dxI_2 = \int_{-k}^{3k} f(x)dxI2​=∫−k3k​f(x)dx, then
  1. (A)I1+2I2=4nkI_1 + 2I_2 = 4nkI1​+2I2​=4nk
  2. (B)I1+2I2=2nkI_1 + 2I_2 = 2nkI1​+2I2​=2nk
  3. (C)I1+nI2=4n2kI_1 + nI_2 = 4n^{2}kI1​+nI2​=4n2k
  4. (D)I1+nI2=6n2kI_1 + nI_2 = 6n^{2}kI1​+nI2​=6n2k

Correct answer: (C)

Step-by-step solution →
Q151·MathematicsSingle correctJEE Main 2022
The value of the integral ∫−22∣x3+x∣(ex∣x∣+1)dx\int_{-2}^{2} \frac{\left|x^{3} + x\right|}{\left(e^{x|x|} + 1\right)}dx∫−22​(ex∣x∣+1)∣x3+x∣​dx is equal to :
  1. (A)5e25e^{2}5e2
  2. (B)3e−23e^{-2}3e−2
  3. (C)4
  4. (D)6

Correct answer: (D)

Step-by-step solution →
Q152·MathematicsSingle correctJEE Main 2022
Let f be a differentiable function in (0,π2)\left(0,\dfrac{\pi}{2}\right)(0,2π​). If ∫cos⁡x1t2f(t)dt=sin⁡3x+cos⁡x\int_{\cos x}^{1}t^{2}f(t)dt=\sin^{3}x+\cos x∫cosx1​t2f(t)dt=sin3x+cosx then 13f′(13)\dfrac{1}{\sqrt{3}}f'\left(\dfrac{1}{\sqrt{3}}\right)3​1​f′(3​1​) is equal to :
  1. (A)6−926-9\sqrt{2}6−92​
  2. (B)6−926-\dfrac{9}{\sqrt{2}}6−2​9​
  3. (C)92−62\dfrac{9}{2}-6\sqrt{2}29​−62​
  4. (D)92−6\dfrac{9}{\sqrt{2}}-62​9​−6

Correct answer: (B)

Step-by-step solution →
Q153·MathematicsSingle correctJEE Main 2022
The integral ∫0117[1x]dx\int_{0}^{1}\dfrac{1}{7^{\left[\frac{1}{x}\right]}}dx∫01​7[x1​]1​dx, where [.] denotes the greatest integer function is equal to
  1. (A)1+6log⁡e(67)1+6\log_{e}\left(\dfrac{6}{7}\right)1+6loge​(76​)
  2. (B)1−6log⁡e(67)1-6\log_{e}\left(\dfrac{6}{7}\right)1−6loge​(76​)
  3. (C)log⁡e(76)\log_{e}\left(\dfrac{7}{6}\right)loge​(67​)
  4. (D)1−7log⁡e(67)1-7\log_{e}\left(\dfrac{6}{7}\right)1−7loge​(76​)

Correct answer: (A)

Step-by-step solution →
Q154·MathematicsNumericalJEE Main 2022
The integral 24π∫02(2−x2)dx(2+x2)4+x4\frac{24}{\pi}\int_{0}^{\sqrt{2}}\frac{\left(2-x^{2}\right)dx}{\left(2+x^{2}\right)\sqrt{4+x^{4}}}π24​∫02​​(2+x2)4+x4​(2−x2)dx​ is equal to ____.

Correct answer: 3

Step-by-step solution →
Q155·MathematicsNumericalJEE Main 2022
Let f(x)=max⁡{∣x+1∣,∣x+2∣,...,∣x+5∣}f(x) = \max\{|x + 1|, |x + 2|, ..., |x + 5|\}f(x)=max{∣x+1∣,∣x+2∣,...,∣x+5∣}. Then ∫−60f(x) dx\int_{-6}^{0} f(x)\,dx∫−60​f(x)dx is equal to ______________.

Correct answer: 21

Step-by-step solution →
Q156·MathematicsNumericalJEE Main 2022
The value of the integral 48π4∫0π(3πx22−x3)sin⁡x1+cos⁡2x dx\frac{48}{\pi^{4}}\int_{0}^{\pi}\left(\frac{3\pi x^{2}}{2} - x^{3}\right)\frac{\sin x}{1 + \cos^{2} x}\,dxπ448​∫0π​(23πx2​−x3)1+cos2xsinx​dx is equal to ________.

Correct answer: 6

Step-by-step solution →
Q157·MathematicsSingle correctJEE Main 2022
The value of ∫0πecos⁡xsin⁡x(1+cos⁡2x)(ecos⁡x+e−cos⁡x)dx\int_0^{\pi} \frac{e^{\cos x}\sin x}{(1+\cos^2 x)(e^{\cos x} + e^{-\cos x})}dx∫0π​(1+cos2x)(ecosx+e−cosx)ecosxsinx​dx is equal to
  1. (A)π24\frac{\pi^2}{4}4π2​
  2. (B)π22\frac{\pi^2}{2}2π2​
  3. (C)π4\frac{\pi}{4}4π​
  4. (D)π2\frac{\pi}{2}2π​

Correct answer: (C)

Step-by-step solution →
Q158·MathematicsNumericalJEE Main 2022
The value of b > 3 for which 12∫3b1(x2−1)(x2−4)dx=log⁡e(4940)12\int_{3}^{b}\frac{1}{\left(x^{2} - 1\right)\left(x^{2} - 4\right)}dx = \log_{e}\left(\frac{49}{40}\right)12∫3b​(x2−1)(x2−4)1​dx=loge​(4049​), is equal to

Correct answer: 6

Step-by-step solution →
Q159·MathematicsSingle correctJEE Main 2022
If bn=∫0π2cos⁡2nxsin⁡xdxb_n = \int_{0}^{\frac{\pi}{2}} \frac{\cos^2 nx}{\sin x} dxbn​=∫02π​​sinxcos2nx​dx, n∈Nn \in Nn∈N, then
  1. (A)b3−b2b_3 - b_2b3​−b2​, b4−b3b_4 - b_3b4​−b3​, b5−b4b_5 - b_4b5​−b4​ are in an A.P. with common difference −2-2−2
  2. (B)1b3−b2,1b4−b3,1b5−b4\frac{1}{b_3 - b_2}, \frac{1}{b_4 - b_3}, \frac{1}{b_5 - b_4}b3​−b2​1​,b4​−b3​1​,b5​−b4​1​ are in an A.P. with common difference 2
  3. (C)b3−b2b_3 - b_2b3​−b2​, b4−b3b_4 - b_3b4​−b3​, b5−b4b_5 - b_4b5​−b4​ are in a G.P.
  4. (D)1b3−b2,1b4−b3,1b5−b4\frac{1}{b_3 - b_2}, \frac{1}{b_4 - b_3}, \frac{1}{b_5 - b_4}b3​−b2​1​,b4​−b3​1​,b5​−b4​1​ are in an A.P. with common difference −2-2−2

Correct answer: (D)

Step-by-step solution →
Q160·MathematicsSingle correctJEE Main 2022
lim⁡n→∞(n2(n2+1)(n+1)+n2(n2+4)(n+2)+n2(n2+9)(n+3)+...+n2(n2+n2)(n+n))\lim_{n\to\infty}\left(\dfrac{n^{2}}{\left(n^{2}+1\right)\left(n+1\right)} + \dfrac{n^{2}}{\left(n^{2}+4\right)\left(n+2\right)} + \dfrac{n^{2}}{\left(n^{2}+9\right)\left(n+3\right)} + ... + \dfrac{n^{2}}{\left(n^{2}+n^{2}\right)\left(n+n\right)}\right)limn→∞​((n2+1)(n+1)n2​+(n2+4)(n+2)n2​+(n2+9)(n+3)n2​+...+(n2+n2)(n+n)n2​) is equal to
  1. (A)π8+14log⁡e2\dfrac{\pi}{8}+\dfrac{1}{4}\log_e 28π​+41​loge​2
  2. (B)π4+18log⁡e2\dfrac{\pi}{4}+\dfrac{1}{8}\log_e 24π​+81​loge​2
  3. (C)π4−18log⁡e2\dfrac{\pi}{4}-\dfrac{1}{8}\log_e 24π​−81​loge​2
  4. (D)π8+log⁡e2\dfrac{\pi}{8}+\log_e \sqrt{2}8π​+loge​2​

Correct answer: (A)

Step-by-step solution →
Q161·MathematicsNumericalJEE Main 2022
Let f(θ)=sin⁡θ+∫−π/2π/2(sin⁡θ+tcos⁡θ)f(t)dtf(\theta)=\sin\theta+\int_{-\pi/2}^{\pi/2}(\sin\theta+t\cos\theta)f(t)dtf(θ)=sinθ+∫−π/2π/2​(sinθ+tcosθ)f(t)dt. Then the value of ∣∫0π/2f(θ)dθ∣\left|\int_{0}^{\pi/2} f(\theta)d\theta\right|​∫0π/2​f(θ)dθ​ is ______ .

Correct answer: 1

Step-by-step solution →
Q162·MathematicsSingle correctJEE Main 2022
The value of the integral ∫−π/2π/2dx(1+ex)(sin⁡6x+cos⁡6x)\displaystyle\int_{-\pi/2}^{\pi/2} \dfrac{dx}{\left(1+e^{x}\right)\left(\sin^{6} x+\cos^{6} x\right)}∫−π/2π/2​(1+ex)(sin6x+cos6x)dx​ is equal to
  1. (A)2π2\pi2π
  2. (B)000
  3. (C)π\piπ
  4. (D)π2\dfrac{\pi}{2}2π​

Correct answer: (C)

Step-by-step solution →
Q163·MathematicsNumericalJEE Main 2022
Let Max0≤x≤2{9−x25−x}=α\underset{0\le x\le 2}{\mathrm{Max}}\left\{\dfrac{9-x^{2}}{5-x}\right\}=\alpha0≤x≤2Max​{5−x9−x2​}=α and Min0≤x≤2{9−x25−x}=β\underset{0\le x\le 2}{\mathrm{Min}}\left\{\dfrac{9-x^{2}}{5-x}\right\}=\beta0≤x≤2Min​{5−x9−x2​}=β If ∫β−832α−1Max{9−x25−x,x}dx=α1+α2log⁡e(815)\displaystyle\int_{\beta-\frac{8}{3}}^{2\alpha-1}\mathrm{Max}\left\{\dfrac{9-x^{2}}{5-x},x\right\}dx=\alpha_{1}+\alpha_{2}\log_{e}\left(\dfrac{8}{15}\right)∫β−38​2α−1​Max{5−x9−x2​,x}dx=α1​+α2​loge​(158​) then α1+α2\alpha_{1}+\alpha_{2}α1​+α2​ is equal to _______

Correct answer: 34

Step-by-step solution →
Q164·MathematicsNumericalJEE Advanced 2021
Let gi:[π8,3π8]→Rg_i : \left[ \frac{\pi}{8}, \frac{3\pi}{8} \right] \to \mathbb{R}gi​:[8π​,83π​]→R, i=1,2i = 1, 2i=1,2, and f:[π8,3π8]→Rf : \left[ \frac{\pi}{8}, \frac{3\pi}{8} \right] \to \mathbb{R}f:[8π​,83π​]→R be functions such that g1(x)=1g_1(x) = 1g1​(x)=1, g2(x)=∣4x−π∣g_2(x) = |4x - \pi|g2​(x)=∣4x−π∣ and f(x)=sin⁡2xf(x) = \sin^2 xf(x)=sin2x, for all x∈[π8,3π8]x \in \left[ \frac{\pi}{8}, \frac{3\pi}{8} \right]x∈[8π​,83π​] Define Si=∫π/83π/8f(x)⋅gi(x) dxS_i = \int_{\pi/8}^{3\pi/8} f(x) \cdot g_i(x) \, dxSi​=∫π/83π/8​f(x)⋅gi​(x)dx, i=1,2i = 1, 2i=1,2 The value of 16S1π\frac{16S_1}{\pi}π16S1​​ is _____.

Correct answer: 2.00

Step-by-step solution →
Q165·MathematicsMultiple correctJEE Advanced 2021
Let f:[−π2,π2]→Rf : \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \to \mathbb{R}f:[−2π​,2π​]→R be a continuous function such that f(0)=1f(0) = 1f(0)=1 and ∫0π3f(t) dt=0\int_0^{\frac{\pi}{3}} f(t)\, dt = 0∫03π​​f(t)dt=0 Then which of the following statements is (are) TRUE?
  1. (A)The equation f(x)−3cos⁡3x=0f(x) - 3\cos 3x = 0f(x)−3cos3x=0 has at least one solution in (0,π3)\left(0, \frac{\pi}{3}\right)(0,3π​)
  2. (B)The equation f(x)−3sin⁡3x=−6πf(x) - 3\sin 3x = -\frac{6}{\pi}f(x)−3sin3x=−π6​ has at least one solution in (0,π3)\left(0, \frac{\pi}{3}\right)(0,3π​)
  3. (C)lim⁡x→0x∫0xf(t)dt1−ex2=−1\lim_{x \to 0} \frac{x\int_0^x f(t)dt}{1 - e^{x^2}} = -1limx→0​1−ex2x∫0x​f(t)dt​=−1
  4. (D)lim⁡x→0sin⁡x∫0xf(t)dtx2=−1\lim_{x \to 0} \frac{\sin x \int_0^x f(t)dt}{x^2} = -1limx→0​x2sinx∫0x​f(t)dt​=−1

Correct answer: (A), (B), (C)

Step-by-step solution →
Q166·MathematicsNumericalJEE Advanced 2021
Let gi:[π8,3π8]→Rg_i : \left[ \frac{\pi}{8}, \frac{3\pi}{8} \right] \to \mathbb{R}gi​:[8π​,83π​]→R, i=1,2i = 1, 2i=1,2, and f:[π8,3π8]→Rf : \left[ \frac{\pi}{8}, \frac{3\pi}{8} \right] \to \mathbb{R}f:[8π​,83π​]→R be functions such that g1(x)=1g_1(x) = 1g1​(x)=1, g2(x)=∣4x−π∣g_2(x) = |4x - \pi|g2​(x)=∣4x−π∣ and f(x)=sin⁡2xf(x) = \sin^2 xf(x)=sin2x, for all x∈[π8,3π8]x \in \left[ \frac{\pi}{8}, \frac{3\pi}{8} \right]x∈[8π​,83π​] Define Si=∫π/83π/8f(x)⋅gi(x) dxS_i = \int_{\pi/8}^{3\pi/8} f(x) \cdot g_i(x) \, dxSi​=∫π/83π/8​f(x)⋅gi​(x)dx, i=1,2i = 1, 2i=1,2 The value of 48S2π2\frac{48S_2}{\pi^2}π248S2​​ is _____.

Correct answer: 1.50

Step-by-step solution →
Q167·MathematicsSingle correctJEE Advanced 2021
Let ψ1:[0,∞)→R\psi_1 : [0, \infty) \to \mathbb{R}ψ1​:[0,∞)→R, ψ2:[0,∞)→R\psi_2 : [0, \infty) \to \mathbb{R}ψ2​:[0,∞)→R, f:[0,∞)→Rf : [0, \infty) \to \mathbb{R}f:[0,∞)→R and g:[0,∞)→Rg : [0, \infty) \to \mathbb{R}g:[0,∞)→R be functions such that f(0)=g(0)=0,f(0) = g(0) = 0,f(0)=g(0)=0, ψ1(x)=e−x+x,x≥0,\psi_1(x) = e^{-x} + x, \quad x \ge 0,ψ1​(x)=e−x+x,x≥0, ψ2(x)=x2−2x−2e−x+2,x≥0,\psi_2(x) = x^2 - 2x - 2e^{-x} + 2, \quad x \ge 0,ψ2​(x)=x2−2x−2e−x+2,x≥0, f(x)=∫−xx(∣t∣−t2)e−t2 dt,x>0f(x) = \int_{-x}^{x} \left( |t| - t^2 \right) e^{-t^2} \, dt, \quad x > 0f(x)=∫−xx​(∣t∣−t2)e−t2dt,x>0 and g(x)=∫0x2t  e−t dt,x>0g(x) = \int_{0}^{x^2} \sqrt{t} \; e^{-t} \, dt, \quad x > 0g(x)=∫0x2​t​e−tdt,x>0 Which of the following statements is TRUE ?
  1. (A)ψ1(x)≤1\psi_1(x) \le 1ψ1​(x)≤1, for all x > 0
  2. (B)ψ2(x)≤0\psi_2(x) \le 0ψ2​(x)≤0, for all x > 0
  3. (C)f(x)≥1−e−x2−23x3+25x5f(x) \ge 1 - e^{-x^2} - \frac{2}{3}x^3 + \frac{2}{5}x^5f(x)≥1−e−x2−32​x3+52​x5, for all x∈(0,12)x \in \left(0, \frac{1}{2}\right)x∈(0,21​)
  4. (D)g(x)≤23x3−25x5+17x7g(x) \le \frac{2}{3}x^3 - \frac{2}{5}x^5 + \frac{1}{7}x^7g(x)≤32​x3−52​x5+71​x7, for all x∈(0,12)x \in \left(0, \frac{1}{2}\right)x∈(0,21​)

Correct answer: (D)

Step-by-step solution →
Q168·MathematicsIntegerJEE Advanced 2021
For any real number x, let [x] denote the largest integer less than or equal to x. If I=∫010[10xx+1]dx,I = \int_0^{10} \left[\sqrt{\frac{10x}{x+1}}\right] dx,I=∫010​[x+110x​​]dx, then the value of 9I is _____.

Correct answer: 182

Step-by-step solution →
Q169·MathematicsSingle correctJEE Main 2021
The function f(x), that satisfies the condition f(x)=x+∫0π/2sin⁡x⋅cos⁡y f(y)dyf(x) = x + \int_{0}^{\pi/2} \sin x \cdot \cos y\, f(y)dyf(x)=x+∫0π/2​sinx⋅cosyf(y)dy, is :
  1. (A)x+23(π−2)sin⁡xx + \frac{2}{3}(\pi - 2)\sin xx+32​(π−2)sinx
  2. (B)x + (π + 2) sinx
  3. (C)x+π2sin⁡xx + \frac{\pi}{2}\sin xx+2π​sinx
  4. (D)x + (π − 2) sinx

Correct answer: (D)

Step-by-step solution →
Q170·MathematicsNumericalJEE Main 2021
If x ϕ(x)=∫5x(3t2−2ϕ′(t))dtx\,\phi(x) = \int_{5}^{x} \left( 3t^{2} - 2\phi'(t) \right) dtxϕ(x)=∫5x​(3t2−2ϕ′(t))dt, x>−2x > -2x>−2, and ϕ(0)=4\phi(0) = 4ϕ(0)=4, then ϕ(2)\phi(2)ϕ(2) is ______.

Correct answer: 4

Step-by-step solution →
Q171·MathematicsSingle correctJEE Main 2021
Let f be a non-negative function in [0,1][0, 1][0,1] and twice differentiable in (0,1)(0, 1)(0,1). If ∫0x1−(f′(t))2 dt=∫0xf(t) dt\int_{0}^{x} \sqrt{1 - \left(f'(t)\right)^{2}}\, dt = \int_{0}^{x} f(t)\, dt∫0x​1−(f′(t))2​dt=∫0x​f(t)dt, 0≤x≤10 \le x \le 10≤x≤1 and f(0)=0f(0) = 0f(0)=0, then lim⁡x→01x2∫0xf(t) dt\lim_{x \to 0} \frac{1}{x^{2}} \int_{0}^{x} f(t)\, dtlimx→0​x21​∫0x​f(t)dt :
  1. (A)equals 0
  2. (B)equals 1
  3. (C)does not exist
  4. (D)equals 12\frac{1}{2}21​

Correct answer: (D)

Step-by-step solution →
Q172·MathematicsSingle correctJEE Main 2021
If [x][x][x] is the greatest integer ≤x\leq x≤x, then π2∫02(sin⁡πx2)(x−[x])[x] dx\pi^2 \int_{0}^{2} \left(\sin \frac{\pi x}{2}\right)(x - [x])^{[x]} \, dxπ2∫02​(sin2πx​)(x−[x])[x]dx is equal to :
  1. (A)2(π−1)2(\pi - 1)2(π−1)
  2. (B)4(π−1)4(\pi - 1)4(π−1)
  3. (C)4(π+1)4(\pi + 1)4(π+1)
  4. (D)2(π+1)2(\pi + 1)2(π+1)

Correct answer: (B)

Step-by-step solution →
Q173·MathematicsNumericalJEE Main 2021
Let [t] denote the greatest integer ≤t\le t≤t. Then the value of 8⋅∫−121([2x]+∣x∣)dx8 \cdot \int_{-\frac{1}{2}}^{1} \left( [2x] + |x| \right) dx8⋅∫−21​1​([2x]+∣x∣)dx is ______.

Correct answer: 5

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Q174·MathematicsSingle correctJEE Main 2021
∫616log⁡ex2log⁡ex2+log⁡e(x2−44x+484)dx\int_{6}^{16} \frac{\log_e x^2}{\log_e x^2 + \log_e\left(x^2 - 44x + 484\right)} dx∫616​loge​x2+loge​(x2−44x+484)loge​x2​dx is equal to:
  1. (A)666
  2. (B)888
  3. (C)555
  4. (D)101010

Correct answer: (C)

Step-by-step solution →
Q175·MathematicsSingle correctJEE Main 2021
The value of the integral ∫01x dx(1+x)(1+3x)(3+x)\int_{0}^{1} \frac{\sqrt{x}\,dx}{(1+x)(1+3x)(3+x)}∫01​(1+x)(1+3x)(3+x)x​dx​ is:
  1. (A)π8(1−32)\frac{\pi}{8}\left(1 - \frac{\sqrt{3}}{2}\right)8π​(1−23​​)
  2. (B)π4(1−36)\frac{\pi}{4}\left(1 - \frac{\sqrt{3}}{6}\right)4π​(1−63​​)
  3. (C)π8(1−36)\frac{\pi}{8}\left(1 - \frac{\sqrt{3}}{6}\right)8π​(1−63​​)
  4. (D)π4(1−32)\frac{\pi}{4}\left(1 - \frac{\sqrt{3}}{2}\right)4π​(1−23​​)

Correct answer: (A)

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Q176·MathematicsSingle correctJEE Main 2021
The value of ∫−1/21/2((x+1x−1)2+(x−1x+1)2−2)1/2dx\int_{-1/\sqrt{2}}^{1/\sqrt{2}} \left(\left(\frac{x+1}{x-1}\right)^2 + \left(\frac{x-1}{x+1}\right)^2 - 2\right)^{1/2} dx∫−1/2​1/2​​((x−1x+1​)2+(x+1x−1​)2−2)1/2dx is:
  1. (A)log⁡e4\log_e 4loge​4
  2. (B)log⁡e16\log_e 16loge​16
  3. (C)2log⁡e162\log_e 162loge​16
  4. (D)4log⁡e(3+22)4\log_e (3 + 2\sqrt{2})4loge​(3+22​)

Correct answer: (B)

Step-by-step solution →
Q177·MathematicsSingle correctJEE Main 2021
If the value of the integral ∫05x+[x]ex−[x]dx=αe−1+β\int_{0}^{5} \frac{x + [x]}{e^{x - [x]}} dx = \alpha e^{-1} + \beta∫05​ex−[x]x+[x]​dx=αe−1+β, where α,β∈R\alpha, \beta \in \mathbf{R}α,β∈R, 5α+6β=05\alpha + 6\beta = 05α+6β=0, and [x][x][x] denotes the greatest integer less than or equal to x; then the value of (α+β)2(\alpha + \beta)^{2}(α+β)2 is equal to :
  1. (A)100100100
  2. (B)252525
  3. (C)161616
  4. (D)363636

Correct answer: (B)

Step-by-step solution →
Q178·MathematicsSingle correctJEE Main 2021
The value of ∫−π2π2(1+sin⁡2x1+πsin⁡x)dx\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \left(\frac{1 + \sin^{2} x}{1 + \pi^{\sin x}}\right) dx∫−2π​2π​​(1+πsinx1+sin2x​)dx is
  1. (A)π2\frac{\pi}{2}2π​
  2. (B)5π4\frac{5\pi}{4}45π​
  3. (C)3π4\frac{3\pi}{4}43π​
  4. (D)3π2\frac{3\pi}{2}23π​

Correct answer: (C)

Step-by-step solution →
Q179·MathematicsSingle correctJEE Main 2021
The value of lim⁡n→∞1n∑r=02n−1n2n2+4r2\lim_{n \to \infty} \frac{1}{n} \sum_{r=0}^{2n-1} \frac{n^2}{n^2 + 4r^2}limn→∞​n1​∑r=02n−1​n2+4r2n2​ is:
  1. (A)12tan⁡−1(2)\frac{1}{2}\tan^{-1}(2)21​tan−1(2)
  2. (B)12tan⁡−1(4)\frac{1}{2}\tan^{-1}(4)21​tan−1(4)
  3. (C)tan⁡−1(4)\tan^{-1}(4)tan−1(4)
  4. (D)14tan⁡−1(4)\frac{1}{4}\tan^{-1}(4)41​tan−1(4)

Correct answer: (B)

Step-by-step solution →
Q180·MathematicsNumericalJEE Main 2021
If ∫0π(sin⁡3x)e−sin⁡2xdx\int_{0}^{\pi}\left(\sin^{3}x\right)e^{-\sin^{2}x}dx∫0π​(sin3x)e−sin2xdx = α − βe∫01t etdt\frac{\beta}{e}\int_{0}^{1}\sqrt{t}\ e^{t}dteβ​∫01​t​ etdt, then α + β is equal to……….

Correct answer: 5

Step-by-step solution →
Q181·MathematicsNumericalJEE Main 2021
Let F:[3,5]→RF:\left[3,5\right] \to RF:[3,5]→R be a twice differentiable function on (3,5)\left(3,5\right)(3,5) such that F(x)=e−x∫3x(3t2+2t+4F′(t))dtF\left(x\right)=e^{-x}\int_3^x\left(3t^2+2t+4F'\left(t\right)\right)dtF(x)=e−x∫3x​(3t2+2t+4F′(t))dt. If F′(4)=αeβ−224(eβ−4)2F'\left(4\right)=\frac{\alpha e^{\beta}-224}{\left(e^{\beta}-4\right)^2}F′(4)=(eβ−4)2αeβ−224​, then α+β\alpha+\betaα+β is equal to ____.

Correct answer: 16

Step-by-step solution →
Q182·MathematicsSingle correctJEE Main 2021
The value of lim⁡n→∞1n∑j=1n(2j−1)+8n(2j−1)+4n\lim_{n \to \infty} \frac{1}{n} \sum_{j=1}^{n} \frac{(2j - 1) + 8n}{(2j - 1) + 4n}limn→∞​n1​∑j=1n​(2j−1)+4n(2j−1)+8n​ is equal to :
  1. (A)2−log⁡e(23)2 - \log_e\left(\frac{2}{3}\right)2−loge​(32​)
  2. (B)1+2log⁡e(32)1 + 2\log_e\left(\frac{3}{2}\right)1+2loge​(23​)
  3. (C)5+log⁡e(32)5 + \log_e\left(\frac{3}{2}\right)5+loge​(23​)
  4. (D)3+2log⁡e(23)3 + 2\log_e\left(\frac{2}{3}\right)3+2loge​(32​)

Correct answer: (B)

Step-by-step solution →
Q183·MathematicsSingle correctJEE Main 2021
The value of the definite integral ∫−π4π4dx(1+excos⁡x)(sin⁡4x+cos⁡4x)\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\frac{dx}{\left(1+e^{x\cos x}\right)\left(\sin^4 x+\cos^4 x\right)}∫−4π​4π​​(1+excosx)(sin4x+cos4x)dx​ is equal to :
  1. (A)−π4-\frac{\pi}{4}−4π​
  2. (B)π22\frac{\pi}{2\sqrt{2}}22​π​
  3. (C)−π2-\frac{\pi}{2}−2π​
  4. (D)π2\frac{\pi}{\sqrt{2}}2​π​

Correct answer: (B)

Step-by-step solution →
Q184·MathematicsNumericalJEE Main 2021
Let the domain of the function f(x)=log⁡4(log⁡5(log⁡3(18x−x2−77)))f\left(x\right)=\log_4\left(\log_5\left(\log_3\left(18x-x^2-77\right)\right)\right)f(x)=log4​(log5​(log3​(18x−x2−77))) be (a,b)\left(a,b\right)(a,b). Then the value of the integral ∫absin⁡3x(sin⁡3x+sin⁡3(a+b−x))dx\int_a^b\frac{\sin^3 x}{\left(\sin^3 x+\sin^3\left(a+b-x\right)\right)}dx∫ab​(sin3x+sin3(a+b−x))sin3x​dx is equal to ____.

Correct answer: 1

Step-by-step solution →
Q185·MathematicsSingle correctJEE Main 2021
The value of the definite integral ∫π/245π/24dx1+tan⁡2x3\int_{\pi/24}^{5\pi/24}\frac{dx}{1+\sqrt[3]{\tan 2x}}∫π/245π/24​1+3tan2x​dx​ is :
  1. (A)π6\frac{\pi}{6}6π​
  2. (B)π12\frac{\pi}{12}12π​
  3. (C)π3\frac{\pi}{3}3π​
  4. (D)π18\frac{\pi}{18}18π​

Correct answer: (B)

Step-by-step solution →
Q186·MathematicsSingle correctJEE Main 2021
The value of the integral ∫−11log⁡(x+x2+1)dx\int_{-1}^{1} \log\left(x + \sqrt{x^2 + 1}\right)dx∫−11​log(x+x2+1​)dx is :
  1. (A)-1
  2. (B)1
  3. (C)2
  4. (D)0

Correct answer: (D)

Step-by-step solution →
Q187·MathematicsSingle correctJEE Main 2021
If [x] denotes the greatest integer less than or equal to x, then the value of the integral ∫−π/2π/2[[x]−sin⁡x]dx\int_{-\pi/2}^{\pi/2}\left[[x]-\sin x\right]dx∫−π/2π/2​[[x]−sinx]dx is equal to :
  1. (A)−π-\pi−π
  2. (B)000
  3. (C)111
  4. (D)π\piπ

Correct answer: (A)

Step-by-step solution →
Q188·MathematicsSingle correctJEE Main 2021
If f:R→Rf:R\to Rf:R→R is given by f(x)=x+1,f(x)=x+1,f(x)=x+1, then the value of lim⁡n→∞1n[f(0)+f(5n)+f(10n)+....+f(5(n−1)n)]\lim_{n\to\infty}\frac{1}{n}\left[f(0)+f\left(\frac{5}{n}\right)+f\left(\frac{10}{n}\right)+....+f\left(\frac{5(n-1)}{n}\right)\right]limn→∞​n1​[f(0)+f(n5​)+f(n10​)+....+f(n5(n−1)​)], is :
  1. (A)32\frac{3}{2}23​
  2. (B)12\frac{1}{2}21​
  3. (C)72\frac{7}{2}27​
  4. (D)52\frac{5}{2}25​

Correct answer: (C)

Step-by-step solution →
Q189·MathematicsSingle correctJEE Main 2021
The value of the integral ∫−11log⁡e(1−x+1+x)dx\int_{-1}^{1} \log_e\left(\sqrt{1-x}+\sqrt{1+x}\right)dx∫−11​loge​(1−x​+1+x​)dx is equal to :
  1. (A)2log⁡e2+π4−12\log_e 2+\frac{\pi}{4}-12loge​2+4π​−1
  2. (B)log⁡e2+π2−1\log_e 2+\frac{\pi}{2}-1loge​2+2π​−1
  3. (C)12log⁡e2+π4−32\frac{1}{2}\log_e 2+\frac{\pi}{4}-\frac{3}{2}21​loge​2+4π​−23​
  4. (D)2log⁡e2+π2−122\log_e 2+\frac{\pi}{2}-\frac{1}{2}2loge​2+2π​−21​

Correct answer: (B)

Step-by-step solution →
Q190·MathematicsSingle correctJEE Main 2021
Let a be a positive real numbers such that ∫0aex−[x]dx=10e−9\int_{0}^{a} e^{x-[x]}dx = 10e - 9∫0a​ex−[x]dx=10e−9 where [x] is the greatest integer less than or equal to x. Then a is equal to :
  1. (A)10+log⁡e210+\log_e 210+loge​2
  2. (B)10−log⁡e(1+e)10-\log_e(1+e)10−loge​(1+e)
  3. (C)10+log⁡e(1+e)10+\log_e(1+e)10+loge​(1+e)
  4. (D)10+log⁡e310+\log_e 310+loge​3

Correct answer: (A)

Step-by-step solution →
Q191·MathematicsSingle correctJEE Main 2021
Let g(t)=∫−π/2π/2cos⁡(π4t+f(x))dx,g(t)=\int_{-\pi/2}^{\pi/2}\cos\left(\frac{\pi}{4}t+f(x)\right)dx,g(t)=∫−π/2π/2​cos(4π​t+f(x))dx, where f(x)=log⁡e(x+x2+1), x∈Rf(x)=\log_{e}\left(x+\sqrt{x^{2}+1}\right),\ x\in Rf(x)=loge​(x+x2+1​), x∈R. Then which one of the following is correct?
  1. (A)g(1)=g(0)g(1)=g(0)g(1)=g(0)
  2. (B)g(1)=2g(0)g(1)=\sqrt{2}g(0)g(1)=2​g(0)
  3. (C)g(1)+g(0)=0g(1)+g(0)=0g(1)+g(0)=0
  4. (D)2g(1)=g(0)\sqrt{2}g(1)=g(0)2​g(1)=g(0)

Correct answer: (D)

Step-by-step solution →
Q192·MathematicsSingle correctJEE Main 2021
Let g(x)=∫0xf(t) dtg(x) = \int_{0}^{x} f(t)\,dtg(x)=∫0x​f(t)dt, where f is continuous function in [0, 3] such that 13≤f(t)≤1\frac{1}{3} \le f(t) \le 131​≤f(t)≤1 for all t ∈ [0, 1] and 0≤f(t)≤120 \le f(t) \le \frac{1}{2}0≤f(t)≤21​ for all t ∈ (1, 3]. The largest possible interval in which g(3) lies is :
  1. (A)[−1,−12]\left[-1, -\frac{1}{2}\right][−1,−21​]
  2. (B)[−32,−1]\left[-\frac{3}{2}, -1\right][−23​,−1]
  3. (C)[13,2]\left[\frac{1}{3}, 2\right][31​,2]
  4. (D)[1, 3]

Correct answer: (C)

Step-by-step solution →
Q193·MathematicsNumericalJEE Main 2021
If [⋅][\cdot][⋅] represents the greatest integer function, then the value of ∣∫0π2[[x2]−cos⁡x]dx∣\left| \int_{0}^{\sqrt{\frac{\pi}{2}}} \left[ \left[x^2\right] - \cos x \right] dx \right|​∫02π​​​[[x2]−cosx]dx​ is _________ .

Correct answer: 1

Step-by-step solution →
Q194·MathematicsSingle correctJEE Main 2021
Let f:R→Rf : R \rightarrow Rf:R→R be defined as f(x)=e−xsin⁡xf(x) = e^{-x}\sin xf(x)=e−xsinx. If F:[0,1]→RF : [0, 1] \rightarrow RF:[0,1]→R is a differentiable function such that F(x)=∫0xf(t) dtF(x) = \int_{0}^{x} f(t)\,dtF(x)=∫0x​f(t)dt, then the value of ∫01(F′(x)+f(x))exdx\int_{0}^{1}\left(F'(x) + f(x)\right)e^{x}dx∫01​(F′(x)+f(x))exdx lies in the interval
  1. (A)[327360,329360]\left[\frac{327}{360}, \frac{329}{360}\right][360327​,360329​]
  2. (B)[330360,331360]\left[\frac{330}{360}, \frac{331}{360}\right][360330​,360331​]
  3. (C)[331360,334360]\left[\frac{331}{360}, \frac{334}{360}\right][360331​,360334​]
  4. (D)[335360,336360]\left[\frac{335}{360}, \frac{336}{360}\right][360335​,360336​]

Correct answer: (B)

Step-by-step solution →
Q195·MathematicsSingle correctJEE Main 2021
If the integral ∫010[sin⁡2πx]ex−[x]dx=αe−1+βe−12+γ\int_{0}^{10}\frac{[\sin 2\pi x]}{e^{x-[x]}}dx = \alpha e^{-1} + \beta e^{-\frac{1}{2}} + \gamma∫010​ex−[x][sin2πx]​dx=αe−1+βe−21​+γ, where α,β,γ\alpha, \beta, \gammaα,β,γ are integers and [x][x][x] denotes the greatest integer less than or equal to xxx, then the value of α+β+γ\alpha + \beta + \gammaα+β+γ is equal to :
  1. (A)000
  2. (B)202020
  3. (C)252525
  4. (D)101010

Correct answer: (A)

Step-by-step solution →
Q196·MathematicsNumericalJEE Main 2021
Let In=∫1ex19(log⁡∣x∣)ndxI_{n} = \int_{1}^{e} x^{19}\left(\log|x|\right)^{n}dxIn​=∫1e​x19(log∣x∣)ndx, where n∈Nn \in Nn∈N. If (20)I10=αI9+βI8(20)I_{10} = \alpha I_{9} + \beta I_{8}(20)I10​=αI9​+βI8​, for natural numbers α\alphaα and β\betaβ, then α−β\alpha - \betaα−β equal to _______.

Correct answer: 1

Step-by-step solution →
Q197·MathematicsNumericalJEE Main 2021
Let f : ℝ → ℝ be a continuous function such that f(x) + f(x + 1) = 2, for all x ∈ ℝ. If I1=∫08f(x)dxI_{1} = \int_{0}^{8} f(x)dxI1​=∫08​f(x)dx and I2=∫−13f(x)dxI_{2} = \int_{-1}^{3} f(x)dxI2​=∫−13​f(x)dx, then the value of I1+2I2I_{1} + 2I_{2}I1​+2I2​ is equal to ________ .

Correct answer: 16

Step-by-step solution →
Q198·MathematicsNumericalJEE Main 2021
Let f : (0, 2) → ℝ be defined as f(x)=log⁡2(1+tan⁡(πx4))f(x) = \log_{2}\left(1+\tan\left(\frac{\pi x}{4}\right)\right)f(x)=log2​(1+tan(4πx​)). Then, lim⁡n→∞2n(f(1n)+f(2n)+....+f(1))\lim_{n \to \infty}\frac{2}{n}\left(f\left(\frac{1}{n}\right)+f\left(\frac{2}{n}\right)+....+f(1)\right)limn→∞​n2​(f(n1​)+f(n2​)+....+f(1)) is equal to _______ .

Correct answer: 1

Step-by-step solution →
Q199·MathematicsSingle correctJEE Main 2021
Consider the integral I=∫010[x] e[x]ex−1dx,I=\int\limits_{0}^{10}\frac{[x]\,e^{[x]}}{e^{x-1}}dx,I=0∫10​ex−1[x]e[x]​dx, where [x] denotes the greatest integer less than or equal to x. Then the value of I is equal to:
  1. (A)9(e−1)9(e-1)9(e−1)
  2. (B)45(e+1)45(e+1)45(e+1)
  3. (C)45(e−1)45(e-1)45(e−1)
  4. (D)9(e+1)9(e+1)9(e+1)

Correct answer: (C)

Step-by-step solution →
Q200·MathematicsSingle correctJEE Main 2021
The value of ∫−π/2π/2cos⁡2x1+3xdx\int_{-\pi/2}^{\pi/2} \frac{\cos^{2}x}{1+3^{x}}dx∫−π/2π/2​1+3xcos2x​dx is:
  1. (A)2π
  2. (B)4π
  3. (C)π2\frac{\pi}{2}2π​
  4. (D)π4\frac{\pi}{4}4π​

Correct answer: (D)

Step-by-step solution →
Q201·MathematicsNumericalJEE Main 2021
The value of the integeral ∫0π∣sin⁡2x∣ dx\int_{0}^{\pi}\left|\sin 2x\right|\,dx∫0π​∣sin2x∣dx is _______

Correct answer: 2

Step-by-step solution →
Q202·MathematicsSingle correctJEE Main 2021
For x>0, if f(x)=∫1xlog⁡et(1+t)dtf(x) = \int_{1}^{x} \frac{\log_{e} t}{(1+t)} dtf(x)=∫1x​(1+t)loge​t​dt, then f(e)+f(1e)f(e) + f\left(\frac{1}{e}\right)f(e)+f(e1​) is equal to :
  1. (A)12\frac{1}{2}21​
  2. (B)−1
  3. (C)1
  4. (D)0

Correct answer: (A)

Step-by-step solution →
Q203·MathematicsSingle correctJEE Main 2021
The value of ∑n=1100∫n−1nex−[x]dx\sum_{n=1}^{100} \int_{n-1}^{n} e^{x-[x]}dx∑n=1100​∫n−1n​ex−[x]dx, where [x] is the greatest integer ≤ x, is:
  1. (A)100 (e–1)
  2. (B)100e
  3. (C)100(1-e)
  4. (D)100 (1 + e)

Correct answer: (A)

Step-by-step solution →
Q204·MathematicsNumericalJEE Main 2021
In Im,n=∫01xm−1(1−x)n−1dxI_{m,n} = \int_{0}^{1} x^{m-1}\left(1-x\right)^{n-1} dxIm,n​=∫01​xm−1(1−x)n−1dx, for m, n ≥\geq≥ 1 and ∫01xm−1+xn−1(1+x)m+ndx=αIm,n\int_{0}^{1}\frac{x^{m-1}+x^{n-1}}{\left(1+x\right)^{m+n}}dx = \alpha I_{m,n}∫01​(1+x)m+nxm−1+xn−1​dx=αIm,n​, α∈R\alpha \in Rα∈R, then α\alphaα equals______________ .

Correct answer: 1

Step-by-step solution →
Q205·MathematicsSingle correctJEE Main 2021
The value of ∫−11x2e[x3]dx\int\limits_{-1}^{1} x^2 e^{[x^3]} dx−1∫1​x2e[x3]dx, where [t][t][t] denotes the greatest integer ≤t\le t≤t, is :
  1. (A)e+13\frac{e+1}{3}3e+1​
  2. (B)e−13e\frac{e-1}{3e}3ee−1​
  3. (C)e+13e\frac{e+1}{3e}3ee+1​
  4. (D)13e\frac{1}{3e}3e1​

Correct answer: (C)

Step-by-step solution →
Q206·MathematicsNumericalJEE Main 2021
The value of ∫3x2^{2}2−3x−6 dx is ______. −2

Correct answer: 19

Step-by-step solution →
Q207·MathematicsSingle correctJEE Main 2021
If In= ∫cotn^{n}n x dx , then: π/4
  1. (A)I2_{2}2​1+I4_{4}4​, I3_{3}3​1+I5_{5}5​ , I4_{4}4​1+I6_{6}6​ are in G.P.
  2. (B)I2_{2}2​1+I4_{4}4​, I3_{3}3​1+I5_{5}5​ , I4_{4}4​1+I6_{6}6​ are in A.P.
  3. (C)I2 + I4, I3 + I5, I4 + I6 are in A.P.
  4. (D)I2 + I4, (I3 + I5)2^{2}2, I4 + I6 are in G.P.

Correct answer: (B)

Step-by-step solution →
Q208·MathematicsSingle correctJEE Main 2021
The value of the integral, ∫13[x2−2x−2]dx,\int_{1}^{3}\left[x^{2}-2x-2\right]dx,∫13​[x2−2x−2]dx, where [x] denotes the greatest integer less than or equal to xxx, is:
  1. (A)−4-4−4
  2. (B)−5-5−5
  3. (C)−2−3−1-\sqrt{2}-\sqrt{3}-1−2​−3​−1
  4. (D)−2−3+1-\sqrt{2}-\sqrt{3}+1−2​−3​+1

Correct answer: (C)

Step-by-step solution →
Q209·MathematicsSingle correctJEE Main 2021
Let f(x)f(x)f(x) be a differentiable function defined on [0,2] such that f′(x)=f′(2−x)f^{'}(x)=f^{'}(2-x)f′(x)=f′(2−x) for all x∈(0,2)x\in(0,2)x∈(0,2), f(0)=1f(0)=1f(0)=1 and f(2)=e2f(2)=\mathrm{e}^{2}f(2)=e2. Then the value of ∫02f(x)dx\int_{0}^{2}f(x)dx∫02​f(x)dx is:
  1. (A)1+e21+e^{2}1+e2
  2. (B)1−e21-e^{2}1−e2
  3. (C)2(1−e2)2(1-e^{2})2(1−e2)
  4. (D)2(1+e2)2(1+e^{2})2(1+e2)

Correct answer: (A)

Step-by-step solution →
Q210·MathematicsNumericalJEE Main 2021
If ∫−aa(∣x∣+∣x−2∣)dx=22\int_{-a}^{a} \left(|x| + |x - 2|\right) dx = 22∫−aa​(∣x∣+∣x−2∣)dx=22 ,(a>2)(a > 2)(a>2) and [x] denotes the greatest integer ≤x\leq x≤x, then ∫a−a(x+[x])dx\int_{a}^{-a} \left(x + [x]\right) dx∫a−a​(x+[x])dx is equal to ______

Correct answer: 3

Step-by-step solution →
Q211·MathematicsNumericalJEE Advanced 2020
Let f:R→Rf : \mathbb{R} \to \mathbb{R}f:R→R be a differentiable function such that its derivative f′f'f′ is continuous and f(π)=−6f(\pi) = -6f(π)=−6. If F:[0,π]→RF : [0,\pi] \to \mathbb{R}F:[0,π]→R is defined by F(x)=∫0xf(t) dtF(x) = \int_{0}^{x} f(t)\,dtF(x)=∫0x​f(t)dt, and if ∫0π(f′(x)+F(x))cos⁡x dx=2\int_{0}^{\pi} (f'(x) + F(x)) \cos x \, dx = 2∫0π​(f′(x)+F(x))cosxdx=2, then the value of f(0)f(0)f(0) is ________

Correct answer: 4.00

Step-by-step solution →
Q212·MathematicsMultiple correctJEE Advanced 2020
Which of the following inequalities is/are TRUE?
  1. (A)∫01xcos⁡x dx≥38\int_{0}^{1} x\cos x\, dx \geq \frac{3}{8}∫01​xcosxdx≥83​
  2. (B)∫01xsin⁡x dx≥310\int_{0}^{1} x\sin x\, dx \geq \frac{3}{10}∫01​xsinxdx≥103​
  3. (C)∫01x2cos⁡x dx≥12\int_{0}^{1} x^{2}\cos x\, dx \geq \frac{1}{2}∫01​x2cosxdx≥21​
  4. (D)∫01x2sin⁡x dx≥29\int_{0}^{1} x^{2}\sin x\, dx \geq \frac{2}{9}∫01​x2sinxdx≥92​

Correct answer: (A), (B), (D)

Step-by-step solution →
Q213·MathematicsSingle correctJEE Main 2020
If I1=∫01(1−x50)100 dxI_1 = \displaystyle\int_0^1 (1-x^{50})^{100}\,dxI1​=∫01​(1−x50)100dx and I2=∫01(1−x50)101 dxI_2 = \displaystyle\int_0^1 (1-x^{50})^{101}\,dxI2​=∫01​(1−x50)101dx such that I2=αI1I_2 = \alpha I_1I2​=αI1​ then α\alphaα equals to:
  1. (A)50495050\dfrac{5049}{5050}50505049​
  2. (B)50505049\dfrac{5050}{5049}50495050​
  3. (C)50505051\dfrac{5050}{5051}50515050​
  4. (D)50515050\dfrac{5051}{5050}50505051​

Correct answer: (C)

Step-by-step solution →
Q214·MathematicsSingle correctJEE Main 2020
The integral ∫12ex⋅x2(2+log⁡ex)dx\int_{1}^{2} e^{x} \cdot x^{2}(2+\log_{e} x)dx∫12​ex⋅x2(2+loge​x)dx equals:
  1. (A)e(4e+1)e(4e + 1)e(4e+1)
  2. (B)4e2−14e^{2} - 14e2−1
  3. (C)e(4e−1)e(4e - 1)e(4e−1)
  4. (D)e(2e−1)e(2e - 1)e(2e−1)

Correct answer: (C)

Step-by-step solution →
Q215·MathematicsSingle correctJEE Main 2020
The value of ∫−π/2π/211+esin⁡xdx\displaystyle\int_{-\pi/2}^{\pi/2} \dfrac{1}{1+e^{\sin x}}dx∫−π/2π/2​1+esinx1​dx is:
  1. (A)3π2\dfrac{3\pi}{2}23π​
  2. (B)π2\dfrac{\pi}{2}2π​
  3. (C)π\piπ
  4. (D)π4\dfrac{\pi}{4}4π​

Correct answer: (B)

Step-by-step solution →
Q216·MathematicsSingle correctJEE Main 2020
Let f(x)=∫x(1+x)2dxf(x) = \int \frac{\sqrt{x}}{(1+x)^{2}} dxf(x)=∫(1+x)2x​​dx (x≥0)(x \ge 0)(x≥0). Then f(3)−f(1)f(3) - f(1)f(3)−f(1) is equal to:
  1. (A)π12+12−34\frac{\pi}{12} + \frac{1}{2} - \frac{\sqrt{3}}{4}12π​+21​−43​​
  2. (B)−π12+12+34-\frac{\pi}{12} + \frac{1}{2} + \frac{\sqrt{3}}{4}−12π​+21​+43​​
  3. (C)π6+12−34\frac{\pi}{6} + \frac{1}{2} - \frac{\sqrt{3}}{4}6π​+21​−43​​
  4. (D)−π6+12+34-\frac{\pi}{6} + \frac{1}{2} + \frac{\sqrt{3}}{4}−6π​+21​+43​​

Correct answer: (A)

Step-by-step solution →
Q217·MathematicsNumericalJEE Main 2020
Let {x} and [x] denote the fractional part of x and the greatest integer ≤x respectively of a real number x. if ∫0n{x}dx\int_{0}^{n} \{x\}dx∫0n​{x}dx, ∫0n[x]dx\int_{0}^{n} [x]dx∫0n​[x]dx and 10(n2−n)10(n^{2} - n)10(n2−n), (n ∈ N, n > 1) are three consecutive terms of a G.P. then n is equal to ……

Correct answer: 21

Step-by-step solution →
Q218·MathematicsSingle correctJEE Main 2020
Let f(x)=∣x−2∣f(x) = |x - 2|f(x)=∣x−2∣ and g(x)=f(f(x))g(x) = f(f(x))g(x)=f(f(x)), x∈[0,4]x \in [0, 4]x∈[0,4]. Then ∫03(g(x)−f(x)) dx\int\limits_{0}^{3} \left( g(x) - f(x) \right)\, dx0∫3​(g(x)−f(x))dx is equal to:
  1. (A)12\frac{1}{2}21​
  2. (B)32\frac{3}{2}23​
  3. (C)111
  4. (D)000

Correct answer: (C)

Step-by-step solution →
Q219·MathematicsSingle correctJEE Main 2020
The integral ∫π/6π/3tan⁡3x⋅sin⁡23x(2sec⁡2x⋅sin⁡23x+3tan⁡x⋅sin⁡6x)dx\int_{\pi/6}^{\pi/3} \tan^{3} x \cdot \sin^{2} 3x \left(2\sec^{2} x \cdot \sin^{2} 3x + 3\tan x \cdot \sin 6x\right) dx∫π/6π/3​tan3x⋅sin23x(2sec2x⋅sin23x+3tanx⋅sin6x)dx is
  1. (A)−118-\frac{1}{18}−181​
  2. (B)−19-\frac{1}{9}−91​
  3. (C)92\frac{9}{2}29​
  4. (D)718\frac{7}{18}187​

Correct answer: (A)

Step-by-step solution →
Q220·MathematicsSingle correctJEE Main 2020
∫−ππ∣π−∣x∣∣dx\int\limits_{-\pi}^{\pi} \left| \pi - |x| \right| dx−π∫π​∣π−∣x∣∣dx is equal to:
  1. (A)π2\pi^{2}π2
  2. (B)π22\frac{\pi^{2}}{2}2π2​
  3. (C)2 π2\sqrt{2}\,\pi^{2}2​π2
  4. (D)2π22\pi^{2}2π2

Correct answer: (A)

Step-by-step solution →
Q221·MathematicsSingle correctJEE Main 2020
If the value of the integral ∫01/2x2(1−x2)3/2dx\int_{0}^{1/2} \frac{x^2}{\left(1 - x^2\right)^{3/2}} dx∫01/2​(1−x2)3/2x2​dx is k6\frac{k}{6}6k​, then kkk is equal to:
  1. (A)23+π2\sqrt{3} + \pi23​+π
  2. (B)32−π3\sqrt{2} - \pi32​−π
  3. (C)23−π2\sqrt{3} - \pi23​−π
  4. (D)32+π3\sqrt{2} + \pi32​+π

Correct answer: (C)

Step-by-step solution →
Q222·MathematicsSingle correctJEE Main 2020
If for all real triplets (a, b, c), f(x)=a+bx+cx2f(x)=a+bx+cx^{2}f(x)=a+bx+cx2; then ∫01f(x) dx\int_{0}^{1} f(x)\,dx∫01​f(x)dx is equal to:
  1. (A)12{f(1)+3f(12)}\dfrac{1}{2}\left\{f(1)+3f\left(\dfrac{1}{2}\right)\right\}21​{f(1)+3f(21​)}
  2. (B)16{f(0)+f(1)+4f(12)}\dfrac{1}{6}\left\{f(0)+f(1)+4f\left(\dfrac{1}{2}\right)\right\}61​{f(0)+f(1)+4f(21​)}
  3. (C)13{f(0)+f(12)}\dfrac{1}{3}\left\{f(0)+f\left(\dfrac{1}{2}\right)\right\}31​{f(0)+f(21​)}
  4. (D)2{3f(1)+2f(12)}2\left\{3f(1)+2f\left(\dfrac{1}{2}\right)\right\}2{3f(1)+2f(21​)}

Correct answer: (B)

Step-by-step solution →
Q223·MathematicsSingle correctJEE Main 2020
The value of ∫02πxsin⁡8xsin⁡8x+cos⁡8x dx\displaystyle\int_{0}^{2\pi} \dfrac{x\sin^{8}x}{\sin^{8}x+\cos^{8}x}\,dx∫02π​sin8x+cos8xxsin8x​dx is equal to:
  1. (A)2π2\pi2π
  2. (B)4π4\pi4π
  3. (C)π2\pi^{2}π2
  4. (D)2π22\pi^{2}2π2

Correct answer: (C)

Step-by-step solution →
Q224·MathematicsSingle correctJEE Main 2020
If I=∫12dx2x3−9x2+12x+4I=\int_{1}^{2}\dfrac{dx}{\sqrt{2x^{3}-9x^{2}+12x+4}}I=∫12​2x3−9x2+12x+4​dx​, then:
  1. (A)116<I2<19\dfrac{1}{16}<I^{2}<\dfrac{1}{9}161​<I2<91​
  2. (B)19<I2<18\dfrac{1}{9}<I^{2}<\dfrac{1}{8}91​<I2<81​
  3. (C)18<I2<14\dfrac{1}{8}<I^{2}<\dfrac{1}{4}81​<I2<41​
  4. (D)16<I2<12\dfrac{1}{6}<I^{2}<\dfrac{1}{2}61​<I2<21​

Correct answer: (B)

Step-by-step solution →
Q225·MathematicsSingle correctJEE Main 2020
The value of α\alphaα for which 4α∫−12e−α∣x∣ dx=54\alpha\int_{-1}^{2} e^{-\alpha|x|}\,dx=54α∫−12​e−α∣x∣dx=5, is
  1. (A)log⁡e(32)\log_e\left(\frac{3}{2}\right)loge​(23​)
  2. (B)log⁡e(43)\log_e\left(\frac{4}{3}\right)loge​(34​)
  3. (C)log⁡e2\log_e\sqrt{2}loge​2​
  4. (D)log⁡e2\log_e 2loge​2

Correct answer: (D)

Step-by-step solution →
Q226·MathematicsMultiple correctJEE Advanced 2019
For a∈Ra \in \mathbb{R}a∈R ∣a∣>1|a| > 1∣a∣>1, let lim⁡n→∞(1+23+....+n3n7/3(1(an+1)2+1(an+2)2+...+1(an+n)2))=54\lim\limits_{n \to \infty}\left( \dfrac{1 + \sqrt[3]{2} + .... + \sqrt[3]{n}}{n^{7/3}\left( \dfrac{1}{(an+1)^{2}} + \dfrac{1}{(an+2)^{2}} + ... + \dfrac{1}{(an+n)^{2}} \right)} \right) = 54n→∞lim​​n7/3((an+1)21​+(an+2)21​+...+(an+n)21​)1+32​+....+3n​​​=54 Then the possible value(s) of a is/are
  1. (A)8
  2. (B)−6-6−6
  3. (C)7
  4. (D)−9-9−9

Correct answer: (A), (D)

Step-by-step solution →
Q227·MathematicsNumericalJEE Advanced 2019
If I=2π∫−π/4π/4dx(1+esin⁡x)(2−cos⁡2x)I = \frac{2}{\pi} \int_{-\pi/4}^{\pi/4} \frac{dx}{\left(1 + e^{\sin x}\right)\left(2 - \cos 2x\right)}I=π2​∫−π/4π/4​(1+esinx)(2−cos2x)dx​ then 27I227 I^227I2 equals ____

Correct answer: 4.00

Step-by-step solution →
Q228·MathematicsNumericalJEE Advanced 2019
The value of the integral ∫0π/23cos⁡θ(cos⁡θ+sin⁡θ)5 dθ\displaystyle\int_{0}^{\pi/2} \dfrac{3\sqrt{\cos\theta}}{\left(\sqrt{\cos\theta} + \sqrt{\sin\theta}\right)^{5}}\, d\theta∫0π/2​(cosθ​+sinθ​)53cosθ​​dθ equals ____

Correct answer: 0.50

Step-by-step solution →
Q229·MathematicsSingle correctJEE Main 2019
If ∫0π/2cot⁡xcot⁡x+cosec x dx=m(π+n)\displaystyle\int_{0}^{\pi/2} \dfrac{\cot x}{\cot x + \text{cosec}\,x}\,dx = m(\pi + n)∫0π/2​cotx+cosecxcotx​dx=m(π+n), then m⋅\cdot⋅n is equal to
  1. (A)1
  2. (B)12\dfrac{1}{2}21​
  3. (C)−12-\dfrac{1}{2}−21​
  4. (D)−1-1−1

Correct answer: (D)

Step-by-step solution →
Q230·MathematicsSingle correctJEE Main 2019
Let f : R →\rightarrow→ R be a continuously differentiable function such that f(2) = 6 and f′(2)=148f'(2) = \frac{1}{48}f′(2)=481​. If ∫6f(x)4t3 dt=(x−2)g(x)\int_{6}^{f(x)} 4t^{3}\, dt = (x-2)g(x)∫6f(x)​4t3dt=(x−2)g(x), then lim⁡x→2g(x)\lim_{x \to 2} g(x)limx→2​g(x) is equal to :
  1. (A)24
  2. (B)18
  3. (C)12
  4. (D)36

Correct answer: (B)

Step-by-step solution →
Q231·MathematicsSingle correctJEE Main 2019
A value of α\alphaα such that ∫αα+1dx(x+α)(x+α+1)=log⁡e(98)\displaystyle\int_{\alpha}^{\alpha+1} \dfrac{dx}{(x+\alpha)(x+\alpha+1)} = \log_{e}\left(\dfrac{9}{8}\right)∫αα+1​(x+α)(x+α+1)dx​=loge​(89​) is:
  1. (A)−12-\dfrac{1}{2}−21​
  2. (B)−2-2−2
  3. (C)12\dfrac{1}{2}21​
  4. (D)2

Correct answer: (B)

Step-by-step solution →
Q232·MathematicsSingle correctJEE Main 2019
The value of ∫02π[sin⁡2x(1+cos⁡3x)] dx\displaystyle\int_{0}^{2\pi} [\sin 2x(1+\cos 3x)]\, dx∫02π​[sin2x(1+cos3x)]dx where [t] denotes the greatest integer function, is:
  1. (A)π\piπ
  2. (B)−2π-2\pi−2π
  3. (C)2π2\pi2π
  4. (D)−π-\pi−π

Correct answer: (D)

Step-by-step solution →
Q233·MathematicsSingle correctJEE Main 2019
lim⁡n→∞((n+1)1/3n4/3+(n+2)1/3n4/3+....+(2n)1/3n4/3)\displaystyle\lim_{n\to\infty}\left(\dfrac{(n+1)^{1/3}}{n^{4/3}}+\dfrac{(n+2)^{1/3}}{n^{4/3}}+....+\dfrac{(2n)^{1/3}}{n^{4/3}}\right)n→∞lim​(n4/3(n+1)1/3​+n4/3(n+2)1/3​+....+n4/3(2n)1/3​) is equal to:
  1. (A)34(2)4/3−34\dfrac{3}{4}(2)^{4/3}-\dfrac{3}{4}43​(2)4/3−43​
  2. (B)43(2)3/4\dfrac{4}{3}(2)^{3/4}34​(2)3/4
  3. (C)34(2)4/3−43\dfrac{3}{4}(2)^{4/3}-\dfrac{4}{3}43​(2)4/3−34​
  4. (D)43(2)4/3\dfrac{4}{3}(2)^{4/3}34​(2)4/3

Correct answer: (A)

Step-by-step solution →
Q234·MathematicsSingle correctJEE Main 2019
The integral ∫π/6π/3sec⁡2/3x cosec⁡4/3x dx\int_{\pi/6}^{\pi/3} \sec^{2/3} x \, \operatorname{cosec}^{4/3} x \, dx∫π/6π/3​sec2/3xcosec4/3xdx is equal to
  1. (A)35/6−32/33^{5/6} - 3^{2/3}35/6−32/3
  2. (B)35/3−31/33^{5/3} - 3^{1/3}35/3−31/3
  3. (C)37/6−35/63^{7/6} - 3^{5/6}37/6−35/6
  4. (D)34/3−31/33^{4/3} - 3^{1/3}34/3−31/3

Correct answer: (C)

Step-by-step solution →
Q235·MathematicsSingle correctJEE Main 2019
The value of ∫0π/2sin⁡3xsin⁡x+cos⁡xdx\displaystyle\int_{0}^{\pi/2} \dfrac{\sin^{3}x}{\sin x+\cos x}dx∫0π/2​sinx+cosxsin3x​dx is:
  1. (A)π−24\dfrac{\pi-2}{4}4π−2​
  2. (B)π−12\dfrac{\pi-1}{2}2π−1​
  3. (C)π−14\dfrac{\pi-1}{4}4π−1​
  4. (D)π−28\dfrac{\pi-2}{8}8π−2​

Correct answer: (C)

Step-by-step solution →
Q236·MathematicsSingle correctJEE Main 2019
The value of the integral ∫01xcot⁡−1(1−x2+x4)dx\displaystyle\int_{0}^{1} x\cot^{-1}\left(1-x^2+x^4\right)dx∫01​xcot−1(1−x2+x4)dx is
  1. (A)π4−12log⁡e2\dfrac{\pi}{4}-\dfrac{1}{2}\log_e 24π​−21​loge​2
  2. (B)π2−log⁡e2\dfrac{\pi}{2}-\log_e 22π​−loge​2
  3. (C)π2−12log⁡e2\dfrac{\pi}{2}-\dfrac{1}{2}\log_e 22π​−21​loge​2
  4. (D)π4−log⁡e2\dfrac{\pi}{4}-\log_e 24π​−loge​2

Correct answer: (A)

Step-by-step solution →
Q237·MathematicsSingle correctJEE Main 2019
If f(x)=2−xcos⁡x2+xcos⁡xf(x)=\frac{2-x\cos x}{2+x\cos x}f(x)=2+xcosx2−xcosx​ and g(x)=log⁡ex,(x>0)g(x)=\log_{e}x,(x>0)g(x)=loge​x,(x>0) then the value of the integral ∫−π/4π/4g(f(x))dx\displaystyle\int_{-\pi/4}^{\pi/4} g\left(f(x)\right)dx∫−π/4π/4​g(f(x))dx is:
  1. (A)log⁡e1\log_{e}1loge​1
  2. (B)log⁡e2\log_{e}2loge​2
  3. (C)log⁡ee\log_{e}eloge​e
  4. (D)log⁡e3\log_{e}3loge​3

Correct answer: (A)

Step-by-step solution →
Q238·MathematicsSingle correctJEE Main 2019
Let f(x)=∫0xg(t) dtf(x)=\int_{0}^{x} g(t)\,dtf(x)=∫0x​g(t)dt, were g is a non zero even function. If f(x+5)=g(x)f(x+5)=g(x)f(x+5)=g(x), then ∫0xf(t) dt\int_{0}^{x} f(t)\,dt∫0x​f(t)dt equals
  1. (A)∫x+55g(t) dt\int_{x+5}^{5} g(t)\,dt∫x+55​g(t)dt
  2. (B)2∫5x−5g(t) dt2\int_{5}^{x-5} g(t)\,dt2∫5x−5​g(t)dt
  3. (C)∫5x+5g(t) dt\int_{5}^{x+5} g(t)\,dt∫5x+5​g(t)dt
  4. (D)5∫x+55g(t) dt5\int_{x+5}^{5} g(t)\,dt5∫x+55​g(t)dt

Correct answer: (A)

Step-by-step solution →
Q239·MathematicsSingle correctJEE Main 2019
The integral ∫1e{(xe)2x−(ex)x}log⁡ex dx\displaystyle\int_{1}^{e} \left\{ \left(\dfrac{x}{e}\right)^{2x} - \left(\dfrac{e}{x}\right)^{x} \right\} \log_{e} x \, dx∫1e​{(ex​)2x−(xe​)x}loge​xdx is equal to :
  1. (A)12−e−1e2\dfrac{1}{2} - e - \dfrac{1}{e^{2}}21​−e−e21​
  2. (B)−12+1e−12e2-\dfrac{1}{2} + \dfrac{1}{e} - \dfrac{1}{2e^{2}}−21​+e1​−2e21​
  3. (C)32−1e−12e2\dfrac{3}{2} - \dfrac{1}{e} - \dfrac{1}{2e^{2}}23​−e1​−2e21​
  4. (D)32−e−12e2\dfrac{3}{2} - e - \dfrac{1}{2e^{2}}23​−e−2e21​

Correct answer: (D)

Step-by-step solution →
Q240·MathematicsSingle correctJEE Main 2019
lim⁡x→∞(nn2+12+nn2+22+nn2+32+…+15n)\lim_{x\to\infty} \left( \dfrac{n}{n^{2}+1^{2}} + \dfrac{n}{n^{2}+2^{2}} + \dfrac{n}{n^{2}+3^{2}} + \ldots + \dfrac{1}{5n} \right)limx→∞​(n2+12n​+n2+22n​+n2+32n​+…+5n1​) is equal to
  1. (A)π4\dfrac{\pi}{4}4π​
  2. (B)tan⁡−1(3)\tan^{-1}(3)tan−1(3)
  3. (C)π2\dfrac{\pi}{2}2π​
  4. (D)tan⁡−1(2)\tan^{-1}(2)tan−1(2)

Correct answer: (D)

Step-by-step solution →
Q241·MathematicsSingle correctJEE Main 2019
Let f and g be continuous functions on [0,a][0, a][0,a] such that f(x)=f(a−x)f(x)=f(a-x)f(x)=f(a−x) and g(x)+g(a−x)=4g(x)+g(a-x)=4g(x)+g(a−x)=4, then ∫0af(x)g(x)dx\int_{0}^{a}f(x)g(x)dx∫0a​f(x)g(x)dx is equal to:
  1. (A)4∫0af(x)dx4\int_{0}^{a}f(x)dx4∫0a​f(x)dx
  2. (B)∫0af(x)dx\int_{0}^{a}f(x)dx∫0a​f(x)dx
  3. (C)2∫0af(x)dx2\int_{0}^{a}f(x)dx2∫0a​f(x)dx
  4. (D)−3∫0af(x)dx-3\int_{0}^{a}f(x)dx−3∫0a​f(x)dx

Correct answer: (C)

Step-by-step solution →
Q242·MathematicsSingle correctJEE Main 2019
The value of the integral ∫−22sin⁡2x[xπ]+12 dx\int_{-2}^{2}\dfrac{\sin^{2}x}{\left[\dfrac{x}{\pi}\right]+\dfrac{1}{2}}\,dx∫−22​[πx​]+21​sin2x​dx (where [x] denotes the greatest integer less than or equal to x) is:
  1. (A)0
  2. (B)sin⁡4\sin 4sin4
  3. (C)4
  4. (D)4−sin⁡44-\sin 44−sin4

Correct answer: (A)

Step-by-step solution →
Q243·MathematicsSingle correctJEE Main 2019
Let I=∫ab(x4−2x2) dxI = \displaystyle\int_{a}^{b} (x^4 - 2x^2)\,dxI=∫ab​(x4−2x2)dx. If I is minimum then the ordered pair (a, b) is:
  1. (A)(0,2)(0, \sqrt{2})(0,2​)
  2. (B)(−2,0)(-\sqrt{2}, 0)(−2​,0)
  3. (C)(2,−2)(\sqrt{2}, -\sqrt{2})(2​,−2​)
  4. (D)(−2,2)(-\sqrt{2}, \sqrt{2})(−2​,2​)

Correct answer: (D)

Step-by-step solution →
Q244·MathematicsSingle correctJEE Main 2019
The value of ∫0π∣cos⁡x∣3 dx\displaystyle\int_{0}^{\pi} |\cos x|^3\, dx∫0π​∣cosx∣3dx is:
  1. (A)0
  2. (B)43\dfrac{4}{3}34​
  3. (C)23\dfrac{2}{3}32​
  4. (D)−43-\dfrac{4}{3}−34​

Correct answer: (B)

Step-by-step solution →
Q245·MathematicsSingle correctJEE Main 2019
If ∫0π3tan⁡θ2ksec⁡θdθ=1−12,(k>0)\int_{0}^{\frac{\pi}{3}} \frac{\tan\theta}{\sqrt{2k\sec\theta}} d\theta = 1 - \frac{1}{\sqrt{2}}, (k > 0)∫03π​​2ksecθ​tanθ​dθ=1−2​1​,(k>0), then the value of k is:
  1. (A)222
  2. (B)12\frac{1}{2}21​
  3. (C)444
  4. (D)111

Correct answer: (A)

Step-by-step solution →
Q246·MathematicsNumericalJEE Advanced 2018
The value of the integral ∫01/21+3((x+1)2(1−x)6)1/4 dx\int_{0}^{1/2} \frac{1+\sqrt{3}}{\left((x+1)^{2}(1-x)^{6}\right)^{1/4}}\, dx∫01/2​((x+1)2(1−x)6)1/41+3​​dx is ______ .

Correct answer: 2

Step-by-step solution →
Q247·MathematicsMultiple correctJEE Advanced 2017
Let f:R→(0,1)f : R \to (0, 1)f:R→(0,1) be a continuous function. Then, which of the following function(s) has(have) the value zero at some point in the interval (0,1)(0, 1)(0,1) ?
  1. (A)ex−∫0xf(t)sin⁡t dte^{x} - \int_{0}^{x} f(t)\sin t\, dtex−∫0x​f(t)sintdt
  2. (B)x9−f(x)x^{9} - f(x)x9−f(x)
  3. (C)f(x)+∫0π/2f(t)sin⁡t dtf(x) + \int_{0}^{\pi/2} f(t)\sin t\, dtf(x)+∫0π/2​f(t)sintdt
  4. (D)x−∫0π2−xf(t)cos⁡t dtx - \int_{0}^{\frac{\pi}{2} - x} f(t)\cos t\, dtx−∫02π​−x​f(t)costdt

Correct answer: (B), (D)

Step-by-step solution →
Q248·MathematicsMultiple correctJEE Advanced 2017
If I=∑k=198∫kk+1k+1x(x+1) dxI = \displaystyle\sum_{k=1}^{98} \int_{k}^{k+1} \dfrac{k+1}{x(x+1)}\,dxI=k=1∑98​∫kk+1​x(x+1)k+1​dx, then
  1. (A)I>log⁡e99I > \log_{e} 99I>loge​99
  2. (B)I<log⁡e99I < \log_{e} 99I<loge​99
  3. (C)I<4950I < \dfrac{49}{50}I<5049​
  4. (D)I>4950I > \dfrac{49}{50}I>5049​

Correct answer: (B), (D)

Step-by-step solution →
Q249·MathematicsIntegerJEE Advanced 2017
Let f:R→Rf : R \to Rf:R→R be a differentiable function such that f(0)=0f(0) = 0f(0)=0, f(π2)=3f\left(\frac{\pi}{2}\right) = 3f(2π​)=3 and f′(0)=1f'(0) = 1f′(0)=1. If g(x)=∫xπ/2[f′(t)cosec⁡t−cot⁡t cosec⁡t f(t)]dtg(x) = \int_{x}^{\pi/2} \left[ f'(t)\operatorname{cosec} t - \cot t\ \operatorname{cosec} t\ f(t) \right] dtg(x)=∫xπ/2​[f′(t)cosect−cott cosect f(t)]dt for x∈(0,π2]x \in \left(0, \frac{\pi}{2}\right]x∈(0,2π​], then lim⁡x→0g(x)=\lim_{x \to 0} g(x) =limx→0​g(x)=

Correct answer: 2

Step-by-step solution →
Q250·MathematicsIntegerJEE Advanced 2016
The total number of distinct x∈[0,1]x \in [0, 1]x∈[0,1] for which ∫0xt21+t4 dt=2x−1\int_{0}^{x} \frac{t^{2}}{1 + t^{4}}\,dt = 2x - 1∫0x​1+t4t2​dt=2x−1 is

Correct answer: 1

Step-by-step solution →
Q251·MathematicsSingle correctJEE Advanced 2016
The value of ∫−π2π2x2cos⁡x1+ex dx\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{x^{2} \cos x}{1 + e^{x}} \, dx∫−2π​2π​​1+exx2cosx​dx is equal to
  1. (A)π24−2\frac{\pi^{2}}{4} - 24π2​−2
  2. (B)π24+2\frac{\pi^{2}}{4} + 24π2​+2
  3. (C)π2−eπ2\pi^{2} - e^{\frac{\pi}{2}}π2−e2π​
  4. (D)π2+eπ2\pi^{2} + e^{\frac{\pi}{2}}π2+e2π​

Correct answer: (A)

Step-by-step solution →
Q252·MathematicsIntegerJEE Advanced 2015
Let f:R→Rf : \mathbb{R} \to \mathbb{R}f:R→R be a function defined by f(x)={[x],x≤20,x>2f(x) = \begin{cases} [x], & x \le 2 \\ 0, & x > 2 \end{cases}f(x)={[x],0,​x≤2x>2​, where [x][x][x] is the greatest integer less than or equal to xxx. If I=∫−12xf(x2)2+f(x+1) dxI = \displaystyle\int_{-1}^{2} \dfrac{x f(x^{2})}{2 + f(x+1)}\, dxI=∫−12​2+f(x+1)xf(x2)​dx, then the value of (4I−1)(4I - 1)(4I−1) is

Correct answer: 0

Step-by-step solution →
Q253·MathematicsIntegerJEE Advanced 2015
Let f:R→Rf : \mathbb{R} \to \mathbb{R}f:R→R be a continuous odd function, which vanishes exactly at one point and f(1)=12f(1) = \dfrac{1}{2}f(1)=21​. Suppose that F(x)=∫−1xf(t) dtF(x) = \displaystyle\int_{-1}^{x} f(t)\,dtF(x)=∫−1x​f(t)dt for all x∈[−1,2]x \in [-1, 2]x∈[−1,2] and G(x)=∫−1xt∣f(f(t))∣dtG(x) = \displaystyle\int_{-1}^{x} t\left|f\left(f(t)\right)\right|dtG(x)=∫−1x​t∣f(f(t))∣dt for all x∈[−1,2]x \in [-1, 2]x∈[−1,2]. If lim⁡x→1F(x)G(x)=114\displaystyle\lim_{x \to 1}\dfrac{F(x)}{G(x)} = \dfrac{1}{14}x→1lim​G(x)F(x)​=141​, then the value of f(12)f\left(\dfrac{1}{2}\right)f(21​) is

Correct answer: 7

Step-by-step solution →
Q254·MathematicsMultiple correctJEE Advanced 2015
The option(s) with the values of aaa and LLL that satisfy the following equation is(are) ∫04πet(sin⁡6at+cos⁡4at)dt∫0πet(sin⁡6at+cos⁡4at)dt=L\dfrac{\displaystyle\int_{0}^{4\pi} e^{t}\left(\sin^{6}at + \cos^{4}at\right)dt}{\displaystyle\int_{0}^{\pi} e^{t}\left(\sin^{6}at + \cos^{4}at\right)dt} = L∫0π​et(sin6at+cos4at)dt∫04π​et(sin6at+cos4at)dt​=L ?
  1. (A)a=2a = 2a=2, L=e4π−1eπ−1L = \dfrac{e^{4\pi} - 1}{e^{\pi} - 1}L=eπ−1e4π−1​
  2. (B)a=2a = 2a=2, L=e4π+1eπ+1L = \dfrac{e^{4\pi} + 1}{e^{\pi} + 1}L=eπ+1e4π+1​
  3. (C)a=4a = 4a=4, L=e4π−1eπ−1L = \dfrac{e^{4\pi} - 1}{e^{\pi} - 1}L=eπ−1e4π−1​
  4. (D)a=4a = 4a=4, L=e4π+1eπ+1L = \dfrac{e^{4\pi} + 1}{e^{\pi} + 1}L=eπ+1e4π+1​

Correct answer: (A), (C)

Step-by-step solution →
Q255·MathematicsMultiple correctJEE Advanced 2015
Let f′(x)=192x32+sin⁡4πxf'(x) = \dfrac{192x^{3}}{2 + \sin^{4}\pi x}f′(x)=2+sin4πx192x3​ for all x∈Rx \in \mathbb{R}x∈R with f(12)=0f\left(\dfrac{1}{2}\right) = 0f(21​)=0. If m≤∫1/21f(x) dx≤Mm \le \displaystyle\int_{1/2}^{1} f(x)\,dx \le Mm≤∫1/21​f(x)dx≤M, then the possible values of mmm and MMM are
  1. (A)m=13m = 13m=13, M=24M = 24M=24
  2. (B)m=14m = \dfrac{1}{4}m=41​, M=12M = \dfrac{1}{2}M=21​
  3. (C)m=−11m = -11m=−11, M=0M = 0M=0
  4. (D)m=1m = 1m=1, M=12M = 12M=12

Correct answer: (D)

Step-by-step solution →
Q256·MathematicsMultiple correctJEE Advanced 2015
Let F:R→RF : \mathbb{R} \to \mathbb{R}F:R→R be a thrice differentiable function. Suppose that F(1)=0F(1) = 0F(1)=0, F(3)=−4F(3) = -4F(3)=−4 and F′(x)<0F'(x) < 0F′(x)<0 for all x∈(1/2,3)x \in (1/2, 3)x∈(1/2,3). Let f(x)=xF(x)f(x) = xF(x)f(x)=xF(x) for all x∈Rx \in \mathbb{R}x∈R. If ∫13x2F′(x) dx=−12\displaystyle\int_{1}^{3} x^{2}F'(x)\,dx = -12∫13​x2F′(x)dx=−12 and ∫13x3F′′(x) dx=40\displaystyle\int_{1}^{3} x^{3}F''(x)\,dx = 40∫13​x3F′′(x)dx=40, then the correct expression(s) is(are)
  1. (A)9f′(3)+f′(1)−32=09f'(3) + f'(1) - 32 = 09f′(3)+f′(1)−32=0
  2. (B)∫13f(x) dx=12\displaystyle\int_{1}^{3} f(x)\,dx = 12∫13​f(x)dx=12
  3. (C)9f′(3)−f′(1)+32=09f'(3) - f'(1) + 32 = 09f′(3)−f′(1)+32=0
  4. (D)∫13f(x) dx=−12\displaystyle\int_{1}^{3} f(x)\,dx = -12∫13​f(x)dx=−12

Correct answer: (C), (D)

Step-by-step solution →
Q257·MathematicsIntegerJEE Advanced 2015
Let F(x)=∫xx2+π62cos⁡2t dtF(x) = \displaystyle\int_{x}^{x^{2}+\frac{\pi}{6}} 2\cos^{2} t\, dtF(x)=∫xx2+6π​​2cos2tdt for all x∈Rx \in \mathbb{R}x∈R and f:[0,12]→[0,∞)f : \left[0, \dfrac{1}{2}\right] \to [0, \infty)f:[0,21​]→[0,∞) be a continuous function. For a∈[0,12]a \in \left[0, \dfrac{1}{2}\right]a∈[0,21​], if F′(a)+2F'(a) + 2F′(a)+2 is the area of the region bounded by x=0x = 0x=0, y=0y = 0y=0, y=f(x)y = f(x)y=f(x) and x=ax = ax=a, then f(0)f(0)f(0) is

Correct answer: 3

Step-by-step solution →
Q258·MathematicsMultiple correctJEE Advanced 2015
Let f(x)=7tan⁡8x+7tan⁡6x−3tan⁡4x−3tan⁡2xf(x) = 7\tan^{8}x + 7\tan^{6}x - 3\tan^{4}x - 3\tan^{2}xf(x)=7tan8x+7tan6x−3tan4x−3tan2x for all x∈(−π2,π2)x \in \left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)x∈(−2π​,2π​). Then the correct expression(s) is(are)
  1. (A)∫0π/4xf(x) dx=112\displaystyle\int_{0}^{\pi/4} x f(x)\,dx = \dfrac{1}{12}∫0π/4​xf(x)dx=121​
  2. (B)∫0π/4f(x) dx=0\displaystyle\int_{0}^{\pi/4} f(x)\,dx = 0∫0π/4​f(x)dx=0
  3. (C)∫0π/4xf(x) dx=16\displaystyle\int_{0}^{\pi/4} x f(x)\,dx = \dfrac{1}{6}∫0π/4​xf(x)dx=61​
  4. (D)∫0π/4f(x) dx=1\displaystyle\int_{0}^{\pi/4} f(x)\,dx = 1∫0π/4​f(x)dx=1

Correct answer: (A), (B)

Step-by-step solution →
Q259·MathematicsIntegerJEE Advanced 2015
If α=∫01(e9x+3tan⁡−1x)(12+9x21+x2)dx\alpha = \displaystyle\int_{0}^{1}\left(e^{9x + 3\tan^{-1}x}\right)\left(\dfrac{12 + 9x^{2}}{1 + x^{2}}\right)dxα=∫01​(e9x+3tan−1x)(1+x212+9x2​)dx where tan⁡−1x\tan^{-1}xtan−1x takes only principal values, then the value of (log⁡e∣1+α∣−3π4)\left(\log_{e}\left|1 + \alpha\right| - \dfrac{3\pi}{4}\right)(loge​∣1+α∣−43π​) is

Correct answer: 9

Step-by-step solution →
Q260·MathematicsMultiple correctJEE Advanced 2014
Let f:(0,∞)→Rf: (0, \infty) \to \mathbb{R}f:(0,∞)→R be given by f(x)=∫1/xxe−(t+1t)dttf(x) = \int_{1/x}^{x} e^{-\left(t+\frac{1}{t}\right)} \frac{dt}{t}f(x)=∫1/xx​e−(t+t1​)tdt​, then
  1. (A)f(x)f(x)f(x) is monotonically increasing on [1,∞)[1, \infty)[1,∞)
  2. (B)f(x)f(x)f(x) is monotonically decreasing on (0,1)(0, 1)(0,1)
  3. (C)f(x)+f(1x)=0f(x) + f\left(\frac{1}{x}\right) = 0f(x)+f(x1​)=0, for all x∈(0,∞)x \in (0, \infty)x∈(0,∞)
  4. (D)f(2x)f(2^x)f(2x) is an odd function of xxx on R\mathbb{R}R

Correct answer: (A), (C), (D)

Step-by-step solution →
Q261·MathematicsSingle correctJEE Advanced 2014
Match the following:
List – IList – II
P.The number of polynomials f(x)f(x)f(x) with non-negative integer coefficients of degree ≤2\le 2≤2, satisfying f(0)=0f(0) = 0f(0)=0 and ∫01f(x) dx=1\int_{0}^{1} f(x)\, dx = 1∫01​f(x)dx=1, is1.8
Q.The number of points in the interval [−13,13]\left[-\sqrt{13}, \sqrt{13}\right][−13​,13​] at which f(x)=sin⁡(x2)+cos⁡(x2)f(x) = \sin(x^{2}) + \cos(x^{2})f(x)=sin(x2)+cos(x2) attains its maximum value, is2.2
R.∫−223x2(1+ex) dx\int_{-2}^{2} \frac{3x^{2}}{(1+e^{x})}\, dx∫−22​(1+ex)3x2​dx equals3.4
S.(∫−1/21/2cos⁡2x⋅log⁡(1+x1−x)dx)(∫01/2cos⁡2x⋅log⁡(1+x1−x)dx)\frac{\left(\int_{-1/2}^{1/2} \cos 2x \cdot \log\left(\frac{1+x}{1-x}\right) dx\right)}{\left(\int_{0}^{1/2} \cos 2x \cdot \log\left(\frac{1+x}{1-x}\right) dx\right)}(∫01/2​cos2x⋅log(1−x1+x​)dx)(∫−1/21/2​cos2x⋅log(1−x1+x​)dx)​ equals4.0
  1. (A)P-3, Q-2, R-4, S-1
  2. (B)P-2, Q-3, R-4, S-1
  3. (C)P-3, Q-2, R-1, S-4
  4. (D)P-2, Q-3, R-1, S-4

Correct answer: (D)

Step-by-step solution →
Q262·MathematicsSingle correctJEE Advanced 2014
Given that for each a∈(0,1)a \in (0, 1)a∈(0,1), lim⁡h→0+∫h1−ht−a(1−t)a−1 dt\lim_{h \to 0^{+}} \int_{h}^{1-h} t^{-a} (1-t)^{a-1}\, dtlimh→0+​∫h1−h​t−a(1−t)a−1dt exists. Let this limit be g(a)g(a)g(a). In addition, it is given that the function g(a)g(a)g(a) is differentiable on (0,1)(0, 1)(0,1). The value of g(12)g\left(\frac{1}{2}\right)g(21​) is
  1. (A)π\piπ
  2. (B)2π2\pi2π
  3. (C)π2\frac{\pi}{2}2π​
  4. (D)π4\frac{\pi}{4}4π​

Correct answer: (A)

Step-by-step solution →
Q263·MathematicsSingle correctJEE Advanced 2014
Let f:[0,2]→Rf : [0, 2] \to \mathbb{R}f:[0,2]→R be a function which is continuous on [0,2][0, 2][0,2] and is differentiable on (0,2)(0, 2)(0,2) with f(0)=1f(0) = 1f(0)=1. Let F(x)=∫0x2f(t)dtF(x) = \int_{0}^{x^{2}} f\left(\sqrt{t}\right) dtF(x)=∫0x2​f(t​)dt for x∈[0,2]x \in [0, 2]x∈[0,2]. If F′(x)=f′(x)F'(x) = f'(x)F′(x)=f′(x) for all x∈(0,2)x \in (0, 2)x∈(0,2), then F(2)F(2)F(2) equals
  1. (A)e2−1e^{2} - 1e2−1
  2. (B)e4−1e^{4} - 1e4−1
  3. (C)e−1e - 1e−1
  4. (D)e4e^{4}e4

Correct answer: (B)

Step-by-step solution →
Q264·MathematicsIntegerJEE Advanced 2014
The value of ∫014x3{d2dx2(1−x2)5}dx\int_{0}^{1} 4x^3 \left\{\frac{d^2}{dx^2}(1 - x^2)^5\right\} dx∫01​4x3{dx2d2​(1−x2)5}dx is __________

Correct answer: 2

Step-by-step solution →
Q265·MathematicsSingle correctJEE Advanced 2014
Given that for each a∈(0,1)a \in (0, 1)a∈(0,1), lim⁡h→0+∫h1−ht−a(1−t)a−1 dt\lim_{h \to 0^{+}} \int_{h}^{1-h} t^{-a} (1-t)^{a-1}\, dtlimh→0+​∫h1−h​t−a(1−t)a−1dt exists. Let this limit be g(a)g(a)g(a). In addition, it is given that the function g(a)g(a)g(a) is differentiable on (0,1)(0, 1)(0,1). The value of g′(12)g'\left(\frac{1}{2}\right)g′(21​) is
  1. (A)π2\frac{\pi}{2}2π​
  2. (B)π\piπ
  3. (C)−π2-\frac{\pi}{2}−2π​
  4. (D)0

Correct answer: (D)

Step-by-step solution →
Q266·MathematicsSingle correctJEE Advanced 2014
The following integral ∫π/4π/2(2cosec⁡x)17 dx\int_{\pi/4}^{\pi/2} (2\operatorname{cosec} x)^{17}\, dx∫π/4π/2​(2cosecx)17dx is equal to
  1. (A)∫0log⁡(1+2)2(eu+e−u)16 du\int_{0}^{\log(1+\sqrt{2})} 2\left(e^{u} + e^{-u}\right)^{16}\, du∫0log(1+2​)​2(eu+e−u)16du
  2. (B)∫0log⁡(1+2)(eu+e−u)17 du\int_{0}^{\log(1+\sqrt{2})} \left(e^{u} + e^{-u}\right)^{17}\, du∫0log(1+2​)​(eu+e−u)17du
  3. (C)∫0log⁡(1+2)(eu−e−u)17 du\int_{0}^{\log(1+\sqrt{2})} \left(e^{u} - e^{-u}\right)^{17}\, du∫0log(1+2​)​(eu−e−u)17du
  4. (D)∫0log⁡(1+2)2(eu−e−u)16 du\int_{0}^{\log(1+\sqrt{2})} 2\left(e^{u} - e^{-u}\right)^{16}\, du∫0log(1+2​)​2(eu−e−u)16du

Correct answer: (A)

Step-by-step solution →
Q267·MathematicsSingle correctJEE Advanced 2013
Let f:[12,1]→Rf:[\frac{1}{2},1]\to Rf:[21​,1]→R (the set of all real numbers) be a positive, non-constant and differentiable function such that f′(x)<2f(x)f'(x)<2f(x)f′(x)<2f(x) and f(12)=1f(\frac{1}{2})=1f(21​)=1. Then the value of ∫1/21f(x) dx\int_{1/2}^{1}f(x)\,dx∫1/21​f(x)dx lies in the interval
  1. (A)(2e−1,2e)(2e-1,2e)(2e−1,2e)
  2. (B)(e−1,2e−1)(e-1,2e-1)(e−1,2e−1)
  3. (C)(e−12,e−1)(\frac{e-1}{2},e-1)(2e−1​,e−1)
  4. (D)(0,e−12)(0,\frac{e-1}{2})(0,2e−1​)

Correct answer: (D)

Step-by-step solution →

Definite Integration — frequently asked

How many questions from Definite Integration appear in JEE?

Definite Integration has appeared in 168 of the last 186 JEE Main and JEE Advanced papers — about 90% of them — contributing 267 questions in total across those papers.

Is Definite Integration an important chapter for JEE?

Judged by how often it is actually tested, it appears in roughly 90% of papers. Chapters above about 50% are effectively guaranteed to show up every session, so they repay thorough preparation; lower-frequency chapters are better treated as targeted revision.

Where do these Definite Integration questions come from?

Every question is from an official JEE Main or JEE Advanced paper, transcribed from the original paper and tagged to this chapter. Answers follow the official answer key.

Other Mathematics chapters

  • Three Dimensional Geometry 344
  • Matrices and Determinants 342
  • Sets, Relations and Functions 313
  • Sequence and Series 287
  • Vector Algebra 245
  • Differential Equations 240
  • Probability 226
  • Permutations and Combinations 220

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